Formulas for Stat 109 Exam 1

525

In a density histogram each rectangular area = percentage (or relative frequency)
1 n
1 k
Mean: y   y i   f i y i   y i p y i 
n i 1
n i 1

SD =
1 n
 yi  y 2 =

n  1 i 1
1 n
2
f i  yi  y  =

n  1 i 1
𝐼𝑓 𝑁 ÷ 4, 𝑊𝑒 𝐴𝑣𝑒𝑟𝑎𝑔𝑒:
𝑛
𝑡ℎ
𝑛
[ ] +[4+1]
𝑄1 = 4
 y
𝑡ℎ
𝑛
4
2
i
2
i
IQR = Q3  Q1
𝐼𝑓 𝑁 𝑖𝑠 𝑛𝑜𝑡 ÷ 4
𝑡ℎ
 x px   
   p yi  =
2
i
step = 1.5 IQR
𝑄1 = [ + 1 ]
2
3𝑛
𝑡ℎ
𝑡ℎ
3𝑛
[ ] +[ 4 +1]
𝑄3 = 4
𝑄3 = [
n 1
~
x
th value
2
𝑡ℎ
3𝑛
+1]
4
2

P A  B  P A  PB  P A  B

P A  B  P A  PB True only when:
A and B are mutually exclusive events.
PB A 
P A  B 
P  A

 y  1sd  contains
 y  2sd  contains
 y  3sd  contains
Y ~ binomial (n, p):
j = 0, 1, 2, …, n
n
n j
PY  j     p j 1  p 
 j

P A  B  P APB A

P A  B  P APB True when A and B are independent events.

Y ~ binomial (n, p): EX    x  np

k  0.5   

PY  k   PY  k  0.5  P Z 





k  0.5   

PY  k   PY  k  0.5  P Z 





k  0.5   
 k  0.5  
PY  k   Pk  0.5  Y  k  0.5  P
Z






Z
Y 

Z
Y 
/ n
68% of the data
95% of the data
99.7% of the data
SDX    x  np1  p 
Z
Y  np
np1  p 
Y 0.5
0.5

 p pˆ 
p
n
n
n
Z

p1  p 
p1  p 
n
n
Table 3
-z
526
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09
-3.4
-3.3
-3.2
-3.1
-3.0
-2.9
-2.8
-2.7
-2.6
-2.5
-2.4
-2.3
-2.2
-2.1
-2.0
-1.9
-1.8
-1.7
-1.6
-1.5
-1.4
-1.3
-1.2
-1.1
-1.0
-0.9
-0.8
-0.7
-0.6
-0.5
-0.4
-0.3
-0.2
-0.1
-0.0
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3.0
3.1
3.2
3.3
3.4
0.000337
0.000483
0.000687
0.000968
0.001350
0.001866
0.002555
0.003467
0.004660
0.006210
0.008198
0.010724
0.013903
0.017864
0.022750
0.028717
0.035930
0.044565
0.054799
0.066807
0.080757
0.096800
0.115070
0.135666
0.158655
0.184060
0.211855
0.241964
0.274253
0.308538
0.344578
0.382089
0.420740
0.460172
0.500000
0.500000
0.539828
0.579260
0.617911
0.655422
0.691462
0.725747
0.758036
0.788145
0.815940
0.841345
0.864334
0.884930
0.903200
0.919243
0.933193
0.945201
0.955435
0.964070
0.971283
0.977250
0.982136
0.986097
0.989276
0.991802
0.993790
0.995340
0.996533
0.997445
0.998134
0.998650
0.999032
0.999313
0.999517
0.999663
0.000325
0.000466
0.000664
0.000935
0.001306
0.001807
0.002477
0.003364
0.004527
0.006037
0.007976
0.010444
0.013553
0.017429
0.022216
0.028067
0.035148
0.043633
0.053699
0.065522
0.079270
0.095098
0.113139
0.133500
0.156248
0.181411
0.208970
0.238852
0.270931
0.305026
0.340903
0.378280
0.416834
0.456205
0.496011
0.503989
0.543795
0.583166
0.621720
0.659097
0.694974
0.729069
0.761148
0.791030
0.818589
0.843752
0.866500
0.886861
0.904902
0.920730
0.934478
0.946301
0.956367
0.964852
0.971933
0.977784
0.982571
0.986447
0.989556
0.992024
0.993963
0.995473
0.996636
0.997523
0.998193
0.998694
0.999065
0.999336
0.999534
0.999675
0.000313
0.000450
0.000641
0.000904
0.001264
0.001750
0.002401
0.003264
0.004396
0.005868
0.007760
0.010170
0.013209
0.017003
0.021692
0.027429
0.034380
0.042716
0.052616
0.064255
0.077804
0.093418
0.111232
0.131357
0.153864
0.178786
0.206108
0.235763
0.267629
0.301532
0.337243
0.374484
0.412936
0.452242
0.492022
0.507978
0.547758
0.587064
0.625516
0.662757
0.698468
0.732371
0.764237
0.793892
0.821214
0.846136
0.868643
0.888768
0.906582
0.922196
0.935745
0.947384
0.957284
0.965620
0.972571
0.978308
0.982997
0.986791
0.989830
0.992240
0.994132
0.995604
0.996736
0.997599
0.998250
0.998736
0.999096
0.999359
0.999550
0.999687
0.000302
0.000434
0.000619
0.000874
0.001223
0.001695
0.002327
0.003167
0.004269
0.005703
0.007549
0.009903
0.012874
0.016586
0.021178
0.026803
0.033625
0.041815
0.051551
0.063008
0.076359
0.091759
0.109349
0.129238
0.151505
0.176186
0.203269
0.232695
0.264347
0.298056
0.333598
0.370700
0.409046
0.448283
0.488034
0.511966
0.551717
0.590954
0.629300
0.666402
0.701944
0.735653
0.767305
0.796731
0.823814
0.848495
0.870762
0.890651
0.908241
0.923641
0.936992
0.948449
0.958185
0.966375
0.973197
0.978822
0.983414
0.987126
0.990097
0.992451
0.994297
0.995731
0.996833
0.997673
0.998305
0.998777
0.999126
0.999381
0.999566
0.999698
0.000291
0.000419
0.000598
0.000845
0.001183
0.001641
0.002256
0.003072
0.004145
0.005543
0.007344
0.009642
0.012545
0.016177
0.020675
0.026190
0.032884
0.040930
0.050503
0.061780
0.074934
0.090123
0.107488
0.127143
0.149170
0.173609
0.200454
0.229650
0.261086
0.294599
0.329969
0.366928
0.405165
0.444330
0.484047
0.515953
0.555670
0.594835
0.633072
0.670031
0.705401
0.738914
0.770350
0.799546
0.826391
0.850830
0.872857
0.892512
0.909877
0.925066
0.938220
0.949497
0.959070
0.967116
0.973810
0.979325
0.983823
0.987455
0.990358
0.992656
0.994457
0.995855
0.996928
0.997744
0.998359
0.998817
0.999155
0.999402
0.999581
0.999709
0.000280
0.000404
0.000577
0.000816
0.001144
0.001589
0.002186
0.002980
0.004025
0.005386
0.007143
0.009387
0.012224
0.015778
0.020182
0.025588
0.032157
0.040059
0.049471
0.060571
0.073529
0.088508
0.105650
0.125072
0.146859
0.171056
0.197663
0.226627
0.257846
0.291160
0.326355
0.363169
0.401294
0.440382
0.480061
0.519939
0.559618
0.598706
0.636831
0.673645
0.708840
0.742154
0.773373
0.802337
0.828944
0.853141
0.874928
0.894350
0.911492
0.926471
0.939429
0.950529
0.959941
0.967843
0.974412
0.979818
0.984222
0.987776
0.990613
0.992857
0.994614
0.995975
0.997020
0.997814
0.998411
0.998856
0.999184
0.999423
0.999596
0.999720
0.000270
0.000390
0.000557
0.000789
0.001107
0.001538
0.002118
0.002890
0.003907
0.005234
0.006947
0.009137
0.011911
0.015386
0.019699
0.024998
0.031443
0.039204
0.048457
0.059380
0.072145
0.086915
0.103835
0.123024
0.144572
0.168528
0.194895
0.223627
0.254627
0.287740
0.322758
0.359424
0.397432
0.436441
0.476078
0.523922
0.563559
0.602568
0.640576
0.677242
0.712260
0.745373
0.776373
0.805105
0.831472
0.855428
0.876976
0.896165
0.913085
0.927855
0.940620
0.951543
0.960796
0.968557
0.975002
0.980301
0.984614
0.988089
0.990863
0.993053
0.994766
0.996093
0.997110
0.997882
0.998462
0.998893
0.999211
0.999443
0.999610
0.999730
0.000260
0.000376
0.000538
0.000762
0.001070
0.001489
0.002052
0.002803
0.003793
0.005085
0.006756
0.008894
0.011604
0.015003
0.019226
0.024419
0.030742
0.038364
0.047460
0.058208
0.070781
0.085343
0.102042
0.121000
0.142310
0.166023
0.192150
0.220650
0.251429
0.284339
0.319178
0.355691
0.393580
0.432505
0.472097
0.527903
0.567495
0.606420
0.644309
0.680822
0.715661
0.748571
0.779350
0.807850
0.833977
0.857690
0.879000
0.897958
0.914657
0.929219
0.941792
0.952540
0.961636
0.969258
0.975581
0.980774
0.984997
0.988396
0.991106
0.993244
0.994915
0.996207
0.997197
0.997948
0.998511
0.998930
0.999238
0.999462
0.999624
0.999740
0.000251
0.000362
0.000519
0.000736
0.001035
0.001441
0.001988
0.002718
0.003681
0.004940
0.006569
0.008656
0.011304
0.014629
0.018763
0.023852
0.030054
0.037538
0.046479
0.057053
0.069437
0.083793
0.100273
0.119000
0.140071
0.163543
0.189430
0.217695
0.248252
0.280957
0.315614
0.351973
0.389739
0.428576
0.468119
0.531881
0.571424
0.610261
0.648027
0.684386
0.719043
0.751748
0.782305
0.810570
0.836457
0.859929
0.881000
0.899727
0.916207
0.930563
0.942947
0.953521
0.962462
0.969946
0.976148
0.981237
0.985371
0.988696
0.991344
0.993431
0.995060
0.996319
0.997282
0.998012
0.998559
0.998965
0.999264
0.999481
0.999638
0.999749
0.000242
0.000349
0.000501
0.000711
0.001001
0.001395
0.001926
0.002635
0.003573
0.004799
0.006387
0.008424
0.011011
0.014262
0.018309
0.023295
0.029379
0.036727
0.045514
0.055917
0.068112
0.082264
0.098525
0.117023
0.137857
0.161087
0.186733
0.214764
0.245097
0.277595
0.312067
0.348268
0.385908
0.424655
0.464144
0.535856
0.575345
0.614092
0.651732
0.687933
0.722405
0.754903
0.785236
0.813267
0.838913
0.862143
0.882977
0.901475
0.917736
0.931888
0.944083
0.954486
0.963273
0.970621
0.976705
0.981691
0.985738
0.988989
0.991576
0.993613
0.995201
0.996427
0.997365
0.998074
0.998605
0.998999
0.999289
0.999499
0.999651
0.999758
+z
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09
Table 3
527
Table 3A:
CDF for the Standard
Normal Distribution
(left tail areas).
-z
-3.4
-3.3
-3.2
-3.1
-3.0
-2.9
-2.8
-2.7
-2.6
-2.5
-2.4
-2.3
-2.2
-2.1
-2.0
-1.9
-1.8
-1.7
-1.6
-1.5
-1.4
-1.3
-1.2
-1.1
-1.0
-0.9
-0.8
-0.7
-0.6
-0.5
-0.4
-0.3
-0.2
-0.1
-0.0
area = Prob[ Z < a ]
a
0
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09
0.000337
0.000483
0.000687
0.000968
0.001350
0.001866
0.002555
0.003467
0.004660
0.006210
0.008198
0.010724
0.013903
0.017864
0.022750
0.028717
0.035930
0.044565
0.054799
0.066807
0.080757
0.096800
0.115070
0.135666
0.158655
0.184060
0.211855
0.241964
0.274253
0.308538
0.344578
0.382089
0.420740
0.460172
0.500000
0.000325
0.000466
0.000664
0.000935
0.001306
0.001807
0.002477
0.003364
0.004527
0.006037
0.007976
0.010444
0.013553
0.017429
0.022216
0.028067
0.035148
0.043633
0.053699
0.065522
0.079270
0.095098
0.113139
0.133500
0.156248
0.181411
0.208970
0.238852
0.270931
0.305026
0.340903
0.378280
0.416834
0.456205
0.496011
0.000313
0.000450
0.000641
0.000904
0.001264
0.001750
0.002401
0.003264
0.004396
0.005868
0.007760
0.010170
0.013209
0.017003
0.021692
0.027429
0.034380
0.042716
0.052616
0.064255
0.077804
0.093418
0.111232
0.131357
0.153864
0.178786
0.206108
0.235763
0.267629
0.301532
0.337243
0.374484
0.412936
0.452242
0.492022
0.000302
0.000434
0.000619
0.000874
0.001223
0.001695
0.002327
0.003167
0.004269
0.005703
0.007549
0.009903
0.012874
0.016586
0.021178
0.026803
0.033625
0.041815
0.051551
0.063008
0.076359
0.091759
0.109349
0.129238
0.151505
0.176186
0.203269
0.232695
0.264347
0.298056
0.333598
0.370700
0.409046
0.448283
0.488034
0.000291
0.000419
0.000598
0.000845
0.001183
0.001641
0.002256
0.003072
0.004145
0.005543
0.007344
0.009642
0.012545
0.016177
0.020675
0.026190
0.032884
0.040930
0.050503
0.061780
0.074934
0.090123
0.107488
0.127143
0.149170
0.173609
0.200454
0.229650
0.261086
0.294599
0.329969
0.366928
0.405165
0.444330
0.484047
0.000280
0.000404
0.000577
0.000816
0.001144
0.001589
0.002186
0.002980
0.004025
0.005386
0.007143
0.009387
0.012224
0.015778
0.020182
0.025588
0.032157
0.040059
0.049471
0.060571
0.073529
0.088508
0.105650
0.125072
0.146859
0.171056
0.197663
0.226627
0.257846
0.291160
0.326355
0.363169
0.401294
0.440382
0.480061
0.000270
0.000390
0.000557
0.000789
0.001107
0.001538
0.002118
0.002890
0.003907
0.005234
0.006947
0.009137
0.011911
0.015386
0.019699
0.024998
0.031443
0.039204
0.048457
0.059380
0.072145
0.086915
0.103835
0.123024
0.144572
0.168528
0.194895
0.223627
0.254627
0.287740
0.322758
0.359424
0.397432
0.436441
0.476078
0.000260
0.000376
0.000538
0.000762
0.001070
0.001489
0.002052
0.002803
0.003793
0.005085
0.006756
0.008894
0.011604
0.015003
0.019226
0.024419
0.030742
0.038364
0.047460
0.058208
0.070781
0.085343
0.102042
0.121000
0.142310
0.166023
0.192150
0.220650
0.251429
0.284339
0.319178
0.355691
0.393580
0.432505
0.472097
0.000251
0.000362
0.000519
0.000736
0.001035
0.001441
0.001988
0.002718
0.003681
0.004940
0.006569
0.008656
0.011304
0.014629
0.018763
0.023852
0.030054
0.037538
0.046479
0.057053
0.069437
0.083793
0.100273
0.119000
0.140071
0.163543
0.189430
0.217695
0.248252
0.280957
0.315614
0.351973
0.389739
0.428576
0.468119
0.000242
0.000349
0.000501
0.000711
0.001001
0.001395
0.001926
0.002635
0.003573
0.004799
0.006387
0.008424
0.011011
0.014262
0.018309
0.023295
0.029379
0.036727
0.045514
0.055917
0.068112
0.082264
0.098525
0.117023
0.137857
0.161087
0.186733
0.214764
0.245097
0.277595
0.312067
0.348268
0.385908
0.424655
0.464144
Table 3
528
Table 3B:
CDF for the Standard
Normal Distribution
(left-tail areas to the right
of the mean).
area = Prob[ Z < a ]
0
+z
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3.0
3.1
3.2
3.3
3.4
.00
0.500000
0.539828
0.579260
0.617911
0.655422
0.691462
0.725747
0.758036
0.788145
0.815940
0.841345
0.864334
0.884930
0.903200
0.919243
0.933193
0.945201
0.955435
0.964070
0.971283
0.977250
0.982136
0.986097
0.989276
0.991802
0.993790
0.995340
0.996533
0.997445
0.998134
0.998650
0.999032
0.999313
0.999517
0.999663
.01
0.503989
0.543795
0.583166
0.621720
0.659097
0.694974
0.729069
0.761148
0.791030
0.818589
0.843752
0.866500
0.886861
0.904902
0.920730
0.934478
0.946301
0.956367
0.964852
0.971933
0.977784
0.982571
0.986447
0.989556
0.992024
0.993963
0.995473
0.996636
0.997523
0.998193
0.998694
0.999065
0.999336
0.999534
0.999675
.02
0.507978
0.547758
0.587064
0.625516
0.662757
0.698468
0.732371
0.764237
0.793892
0.821214
0.846136
0.868643
0.888768
0.906582
0.922196
0.935745
0.947384
0.957284
0.965620
0.972571
0.978308
0.982997
0.986791
0.989830
0.992240
0.994132
0.995604
0.996736
0.997599
0.998250
0.998736
0.999096
0.999359
0.999550
0.999687
.03
0.511966
0.551717
0.590954
0.629300
0.666402
0.701944
0.735653
0.767305
0.796731
0.823814
0.848495
0.870762
0.890651
0.908241
0.923641
0.936992
0.948449
0.958185
0.966375
0.973197
0.978822
0.983414
0.987126
0.990097
0.992451
0.994297
0.995731
0.996833
0.997673
0.998305
0.998777
0.999126
0.999381
0.999566
0.999698
.04
0.515953
0.555670
0.594835
0.633072
0.670031
0.705401
0.738914
0.770350
0.799546
0.826391
0.850830
0.872857
0.892512
0.909877
0.925066
0.938220
0.949497
0.959070
0.967116
0.973810
0.979325
0.983823
0.987455
0.990358
0.992656
0.994457
0.995855
0.996928
0.997744
0.998359
0.998817
0.999155
0.999402
0.999581
0.999709
.05
0.519939
0.559618
0.598706
0.636831
0.673645
0.708840
0.742154
0.773373
0.802337
0.828944
0.853141
0.874928
0.894350
0.911492
0.926471
0.939429
0.950529
0.959941
0.967843
0.974412
0.979818
0.984222
0.987776
0.990613
0.992857
0.994614
0.995975
0.997020
0.997814
0.998411
0.998856
0.999184
0.999423
0.999596
0.999720
a
.06
0.523922
0.563559
0.602568
0.640576
0.677242
0.712260
0.745373
0.776373
0.805105
0.831472
0.855428
0.876976
0.896165
0.913085
0.927855
0.940620
0.951543
0.960796
0.968557
0.975002
0.980301
0.984614
0.988089
0.990863
0.993053
0.994766
0.996093
0.997110
0.997882
0.998462
0.998893
0.999211
0.999443
0.999610
0.999730
.07
0.527903
0.567495
0.606420
0.644309
0.680822
0.715661
0.748571
0.779350
0.807850
0.833977
0.857690
0.879000
0.897958
0.914657
0.929219
0.941792
0.952540
0.961636
0.969258
0.975581
0.980774
0.984997
0.988396
0.991106
0.993244
0.994915
0.996207
0.997197
0.997948
0.998511
0.998930
0.999238
0.999462
0.999624
0.999740
.08
0.531881
0.571424
0.610261
0.648027
0.684386
0.719043
0.751748
0.782305
0.810570
0.836457
0.859929
0.881000
0.899727
0.916207
0.930563
0.942947
0.953521
0.962462
0.969946
0.976148
0.981237
0.985371
0.988696
0.991344
0.993431
0.995060
0.996319
0.997282
0.998012
0.998559
0.998965
0.999264
0.999481
0.999638
0.999749
.09
0.535856
0.575345
0.614092
0.651732
0.687933
0.722405
0.754903
0.785236
0.813267
0.838913
0.862143
0.882977
0.901475
0.917736
0.931888
0.944083
0.954486
0.963273
0.970621
0.976705
0.981691
0.985738
0.988989
0.991576
0.993613
0.995201
0.996427
0.997365
0.998074
0.998605
0.998999
0.999289
0.999499
0.999651
0.999758
Stat 109
Sample Exam 1A
Name_____________
1.)
Given a scenario where 30% of the salmon in a river are
farm fish, and the rest are wild, and we know that 12% the wild
salmon are tagged while 85% of the farm fish are tagged, find the
following probabilities:
Declare all the event
space variables: 2pts
Answer all questions first with complete probability notation.
Then answer each question with an English sentence.
a) Find the probability that a randomly drawn salmon is tagged.
7pts
b) Find the probability of drawing a wild salmon given that the fish is tagged. 7pts
c) Are the type of salmon drawn and whether it is tagged or not independent events? 5pts
(Again use probability notation and numerical values to support your answer.)
529
Stat 109
2.)
Sample Exam 1A
530
John checks his chicken coop each morning and finds the following
number of eggs according to the probability distribution table.
2a.)
Find the expected number of eggs (the mean of x.)
5 pts
x, eggs
2
3
4
p(x)
0.3
0.5
0.2
2b.) Find the typical range of eggs collected each morning expressed as the first standard
deviation of x. (show your work.) 8pts
3.)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Data display for year 2000: widowed females per 1000 for all 58 California counties.
57
59
66
76
78
80
80
81
82
83
84
84
85
86
87
88
89
90
90
90
MONO
SAN BENITO
ALPINE
YOLO
SANTA CLARA
EL DORADO
MADERA
SAN BERNARDINO
VENTURA
ORANGE
KINGS
SANTA CRUZ
SAN DIEGO
SOLANO
LOS ANGELES
ALAMEDA
MONTEREY
KERN
LASSEN
PLACER
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
91
91
91
92
93
93
93
93
94
97
97
97
98
98
99
100
102
102
103
104
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
FRESNO
MERCED
TULARE
CONTRA COSTA
MARIN
PLUMAS
SANTA BARBARA
STANISLAUS
SACRAMENTO
SAN JOAQUIN
SAN MATEO
YUBA
MENDOCINO
RIVERSIDE
SONOMA
GLENN
IMPERIAL
SAN LUIS OBISPO
SAN FRANCISCO
COLUSA
104
104
105
106
109
111
112
113
116
116
119
120
121
123
125
135
138
146
HUMBOLDT
NEVADA
SUTTER
DEL NORTE
SIERRA
CALAVERAS
SHASTA
TUOLUMNE
MARIPOSA
SISKIYOU
BUTTE
TEHAMA
NAPA
TRINITY
AMADOR
INYO
LAKE
MODOC
3a.) Is the 200 California county
widowed data symmetrical or is it
skewed to left or right?
2pts
Frequency
2000 CA Widowed Females per 1000 by County
18
16
14
12
10
8
6
4
2
0
60
80
100
120
140
Stat 109
Sample Exam 1A
531
3b.) Use correct notation to express the 5 key values for a boxplot for the incidence of widowed
females per 1000 for all 58 California counties:
3c.) Draw a boxplot of the CA county widow data set. Properly denote any outliers in the plot.
11pts











Stat 109
Sample Exam 1A
4.)
The lengths of adult Koi in a large pond are normally
distributed with a mean length of 25 inches and a standard
deviation of 3 inches. Answer the following Questions using
the event variable with probability notation and show the
equivalent Z score within probability notation.
532
a)
Declare the event variable. 1pt
4a.)
What probability corresponds to a Koi that is at most 30 inches in length? 8pts
4b.)
The top 29% of Koi lengths must be above what length threshold? 8pts
4c.) How long will a Koi’s length be if it’s length is at the 37th percentile of length? 8pts
4d.) The Koi population of this particular pond is the progeny of a
metallic “Ogon” variety such that each of the Koi presents a number of
metallic scales on its body interspersed among the other scales of the
fish body. Assuming that the number of metallic scales on any given
Koi are normally distributed throughout the population, with a mean of
18 metallic scales and a standard deviation of 3 metallic scales,
determine the following probabilities.
a)
Declare the event variable. 1pt
4e.) Find the probability that a randomly drawn Koi has less than 16 metallic scales upon its body. 8pt
4f.) Find the probability that a randomly drawn Koi has more than 17 metallic scales upon its body. 8pt
4g.) Find the probability that a randomly drawn Koi has between 13 and 19 (inclusive) metallic scales
upon its body. 8pt
Stat 109
Sample Exam 1A
533
5a.) In 2005 57% of the incoming Freshman class for the California State University required either
English or Mathematic remediation of before taking college level course work. 47% required
remediation in English while 37% required remediation in Math. What portion of the 2005 Freshman
class required remediation in both English and Math? Use probability notation to express your
answer. 8pts
5b.) What is the probability that the a 2005 Freshman needs remediation in Math but not English?
Use probability notation to express your answer. 8pts
5c.) Find the probability that either 6 or 7 of 10 randomly drawn CSU Freshman will not require
remediation class work. Declare the variable. Use probability notation to express your answer. 8pts
5d.)
6.)
Find the first standard deviation window to express the typical range of 10 randomly chosen
Freshman that will not require remediation?
8pts
Given that one out of four salamanders are consumed as prey before reaching sexual maturity,
Find the probability that between 10% and 30% of 50 juvenile salamanders will be consumed
before reaching maturity.
Show all work and use proper notation for full credit. Finish with an English sentence.
12pts
Stat 109
Sample Exam 1A KEY
534
1.)
Given a scenario where 30% of the salmon in a river are
farm fish, and the rest are wild, and we know that 12% the wild
salmon are tagged while 85% of the farm fish are tagged, find the
following probabilities:
Answer all questions first with complete probability notation.
Then answer each question with an English sentence.
a)
Declare all the event
space variables: 2pts
T = Tagged Salmon
W = Wild Salmon
F = Farm Salmon
Find the probability that a randomly drawn salmon is tagged.
PT   PT W  PW   PT F  PF 
PT   0.12  0.7  0.85  0.3
PT   .339
b)
7pts
Notation: 2pts
Calculation: 5pts
“About 34% of random draws will be tagged.”
Find the probability of drawing a wild salmon given that the fish is tagged.
7pts
PW T  
PT W  PW 
PW  T 

PT 
PT 
0.12  0.70
P W T  
= 0.2478
0.339
Notation: 2pts
Calculation: 5pts
“About 25% of tagged fish will be wild.”
c) Are the type of salmon drawn and whether it is tagged or not independent events? 5pts
(Again use probability notation and numerical values to support your answer.)
For independent events the following must be true: P A  B  P A PB
We will use the tagged and wild salmon data since it is at hand:
PW  T   PW   PT  ? Is this true?
???
PW  T   PT W   PW  from above
PW T   0.12  0.7
PW T   0.084
Since 0.084  0.2373
Then: PW  T   PW  PT 
Therefore we do not have independent
events.
PW  PT   0.7  0.339
PW  PT   0.2373
In English:
“The probability of whether a salmon is
tagged or not depends on whether it is a
wild or a farmed salmon.”
Stat 109
Sample Exam 1A KEY
535
2.) John checks his chicken coop each morning and finds the following number of eggs
according to the probability distribution table. Express answer with correct notation.
2a.)
Find the expected number of eggs (the mean of x.)
   xi pxi   2  0.3  3  0.5  4  0.2 = 2.9

5 pts
  2.9
x, eggs
2
3
4
1
pt for Notation
2
2b.) Find the typical range of eggs collected each morning expressed as the first standard
deviation of x. (show your work.) 8pts (1 pt for notation)

 x px   
2
i
i
2
 2 2 .3  3 2 .5  4 2 .2  2.9 2  0.49  0.7 
  0.7
𝝁 ± 𝝈 = 𝟐. 𝟗 ± 𝟎. 𝟕 = (𝟐. 𝟐, 𝟑. 𝟔)
3.)
Data display for year 2000: Widowed
females per 1000 for all 58 California counties.
57
59
66
76
78
80
80
81
82
83
84
84
85
86
87
88
89
90
90
90
MONO
SAN BENITO
ALPINE
YOLO
SANTA CLARA
EL DORADO
MADERA
SAN BERNARDINO
VENTURA
ORANGE
KINGS
SANTA CRUZ
SAN DIEGO
SOLANO
LOS ANGELES
ALAMEDA
MONTEREY
KERN
LASSEN
PLACER
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
91
91
91
92
93
93
93
93
94
97
97
97
98
98
99
100
102
102
103
104
FRESNO
MERCED
TULARE
CONTRA COSTA
MARIN
PLUMAS
SANTA BARBARA
STANISLAUS
SACRAMENTO
SAN JOAQUIN
SAN MATEO
YUBA
MENDOCINO
RIVERSIDE
SONOMA
GLENN
IMPERIAL
SAN LUIS OBISPO
SAN FRANCISCO
COLUSA
3a.) Is the 200 California county widowed data
symmetrical or is it skewed to left or right?
Skewed slightly right 2pts
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
104
104
105
106
109
111
112
113
116
116
119
120
121
123
125
135
138
146
HUMBOLDT
NEVADA
SUTTER
DEL NORTE
SIERRA
CALAVERAS
SHASTA
TUOLUMNE
MARIPOSA
SISKIYOU
BUTTE
TEHAMA
NAPA
TRINITY
AMADOR
INYO
LAKE
MODOC
2000 CA Widowed Females per 1000 by County
Frequency
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
18
16
14
12
10
8
6
4
2
0
60
80
100
120
140
p(x)
0.3
0.5
0.2
Stat 109
Sample Exam 1A KEY
536
3b.) Use correct notation to express the 5 key values for a boxplot in correct notation
for the incidence of widowed females per 1000 for all 58 California counties:
11 pts
2.) Find the median:
3.) Determine the Quartile Criterion
𝑛 + 1 𝑡ℎ
𝑥̃ = (
)
2
𝐼𝑓 𝑁 ÷ 4, 𝑊𝑒 𝐴𝑣𝑒𝑟𝑎𝑔𝑒:
𝐼𝑓 𝑁 𝑖𝑠 𝑛𝑜𝑡 ÷ 4
𝑡ℎ
𝑛 𝑡ℎ
𝑛
[4] +[4+1]
𝑄1
58 + 1 𝑡ℎ
=(
)
2
[
2
𝑡ℎ
3𝑛 𝑡ℎ
3𝑛
[ 4 ] +[ 4 +1]
2
𝑄3
59 𝑡ℎ
=(
)
2
𝑡ℎ
𝑛
+1]
4
𝑡ℎ
3𝑛
[
+1]
4
= 29.5𝑡ℎ

= 29𝑡ℎ + 0.5(30𝑡ℎ − 29𝑡ℎ )

Where square brackets indicate that we round
any decimal down to find the nth value in the
data set.
N = 58 is not divisible by 4 so:

𝑄1 = [ 4 + 1 ] 𝑎𝑛𝑑 𝑄3 = [
= 94 + 0.5(97 − 94)
= 94 + 0.5(3)
𝑥̃ = 95.5
4.) Find the 1st and 3rd Quartiles.
𝑄1 = [
=[
𝑡ℎ
𝑛
+1]
4
58
+1]
4
=[
𝑡ℎ
3 ∙ 58
+1]
4
= [ 14.5 + 1 ]𝑡ℎ
= [ 43.5 + 1 ]𝑡ℎ
= [ 15.5 ]𝑡ℎ
= [ 44.5 ]𝑡ℎ
= 15𝑡ℎ 𝑣𝑎𝑙𝑢𝑒
= 44
𝑄1 = 87
𝑄3 = 106
𝑡ℎ
𝑡ℎ
3𝑛
4
+1]
𝑡ℎ
5.) Find the IQR and Step.
𝑡ℎ
3𝑛
𝑄3 = [
+1]
4
𝑡ℎ
𝑛
𝑣𝑎𝑙𝑢𝑒
Outliers:
LOT = 𝑄1 − 𝑆𝑡𝑒𝑝
UOT = 𝑄3 + 𝑆𝑡𝑒𝑝
LOT = 87 − 28.5
UOT = 106 + 28.5
LOT = 58.5
UOT = 134.5
IQR = Inter-quartile range
IQR = Q3 – Q1
IQR = 106 – 87
IQR = 19
𝑆𝑡𝑒𝑝 = 1.5 × 𝐼𝑄𝑅
𝑆𝑡𝑒𝑝 = 1.5 × 19
𝑆𝑡𝑒𝑝 = 28.5
Stat 109
Sample Exam 1A KEY
537
3c.) Draw a boxplot of the CA county widow data set. Properly denote any outliers in the plot.
Note! All data points that lie outside the outlier thresholds are expressed with
asterisks. The boxplot whiskers terminate at the last data point to lie within the
outlier thresholds. It is an error to extend the whiskers to the outlier thresholds.
o

o







4.)
The lengths of adult Koi in a large pond are normally distributed
with a mean length of 25 inches and a standard deviation of 3 inches.
Answer the following Questions using the event variable with probability
notation and show the equivalent Z score within probability notation.

o
o


Declare the event variable.
X = Koi length in inches. 1pt
X = Fish. -1pt: For a vague
declaration without units of measure
4a.)
4a.) What probability corresponds to a Koi at most 30 inches in length? 8pts Note! Here is a number with a
circle drawn around it:
X 
30  25
 Z 
Z = 1.67, then
Z

3
0.95254 -1pt
P X  30  PZ  1.67  0.95254
This number has no meaning
without its supporting notation.
Answer like this with complete notation, Not like this:
4b.) The top 29% of Koi lengths must be above what length threshold? 8pts
Note: This is an upper tail probability and we must know
to use the compliment value because the normal distribution
tables only provide for lower tail probabilities.
P X  ???  PZ  ???  0.29
a)
b)
c)
d)
Search the body of the table for Z score associated with 0.71
Then PZ  0.55  0.29
X 
X  25
Find the X associated with this Z score: Z 
 0.55 
 X = 26.65

3
Report the answer using probability notation.
Show the transition from the real world X values to the associated Z scores.
P X  26.65  PZ  0.55  0.29
Stat 109
Sample Exam 1A KEY
538
4c.) How long will a Koi’s length be if it’s length is at
the 37th percentile of length? 8pts
Note that the percentiles on the normal
curve run from zero to 100% and
accumulate from left to right.
0%
10%
20%
22"
The corresponding Koi lengths for each
of these percentiles are shown below.
30%
23"
40%
24"
50%
60%
70%
80%
25"
26"
27"
28"
90%
100%
c) We find:
d) Using Standardization we find x:
x
x  25
PZ  0.33  0.37
Z
 0.33 

3
e) Report your answer with full probability notation:
a) Search the body of the table for Z
score associated with 0.37.
P X  24  PZ  0.33  0.37
b) P X  ???   PZ  ???   0.37
4d.) The Koi population of this particular pond is the progeny of a
a) Declare the event variable. 1pt
metallic “Ogon” variety such that each of the Koi presents a number of N = The number of metallic
metallic scales on its body interspersed among the other scales of the
scales found on given Koi
fish body. Assuming that the number of metallic scales on any given
from the pond.
Koi are normally distributed throughout the population, with a mean of
N = Scales -1pt:
18 metallic scales and a standard deviation of 3 metallic scales,
For a vague declaration
determine the following probabilities.
4e.) Find the probability that a randomly drawn Koi has less than 16 metallic scales upon its body. 8pt
PN  16  PN  15
Note that discrete counts must be adjusted for a mapping to the continuous
normal curve. Strict inequalities of “less than” must be converted
numerically to “less than or equal to” if using a continuous distribution.
PN  16  PN  15  PN  15.5
Z
x


15.5  18
3
A continuity correction is necessary here because a
discrete count of fish scales must be mapped to a
continuous curve of the normal distribution. Note that we
always choose a half count in the direction that enlarges
the shaded region beneath the normal curve.
Z   0 .8 3
PN  16  PZ  0.83  0.203269
In English: “About 20% of the time a
randomly drawn Koi will have less than 16
metallic scales upon its body.
Answer with complete notation like this, and not this:
0.203269
Stat 109
Sample Exam 1A KEY
539
4f.) Find the probability that a randomly drawn Koi has more than 17 metallic scales upon its body. 8pt
PN  17  PN  18
Note that discrete counts must be adjusted for a mapping to the continuous
normal curve. Strict inequalities of “greater than” must be converted
numerically to “greater than or equal to” if using a continuous distribution.
PN  17  PN  18  PN  17.5
Z
x


A continuity correction is necessary here because a
discrete count of fish scales must be mapped to a
continuous curve of the normal distribution. Note that we
always choose a half count in the direction that enlarges
the shaded region beneath the normal curve. Also note
that a compliment of one minus the Z-table probability
must be accounted for here.
17.5  18
3
Z  0 .1 6
PZ  0.17  1  PZ  0.17  1  0.432505
PN  17  PZ  0.17  0.567495
Answer with complete notation like this, and not this:
In English: “About 57% of the time a
randomly drawn Koi will have more than 17
metallic scales upon its body.
0.567495
4g.) Find the probability that a randomly drawn Koi has between 13 and 19 (inclusive) metallic scales
upon its body. 8pt
P13  N  19
Note that discrete counts must be adjusted for a mapping to the continuous normal
curve. Here the inclusive counts (of 13 and 19) are already specified in the problem.
P13  N  19  P12.5  N  19.5
A continuity correction is still necessary because a discrete count of fish scales must be mapped to a
continuous curve of the normal distribution. Note that we always choose a half count in the direction that
enlarges the shaded region beneath the normal curve.
Z
x

Z  1 .8 3

12.5  18
3
Z
x

Z  0.5

19.5  18
3
Stat 109
Sample Exam 1A KEY
540
4g.) Find the probability that a randomly drawn Koi has between 13 and 19 (inclusive) metallic scales
upon its body. 8pt Continued
P13  N  19  P 1.83  Z  0.5
P13  N  19  PZ  0.5  PZ  1.83
P13  N  19  0.691462  0.033625
P13  N  19  0.657837
In English: “About 66% of the time a randomly
drawn Koi will have between 13 and 19 metallic
scales upon its body.
Answer with complete notation
like this, and not this
P13  N  19  P 1.83  Z  0.5  0.657837
0.657837
5a.) In 2005 57% of the incoming Freshman class for the California State University required either
English or Mathematic remediation of before taking college level course work. 47% required remediation in
English while 37% required remediation in Math. What portion of the 2005 Freshman class required
remediation in both English and Math? Use probability notation to express your answer. 8pts
PE  M   PE   PM   PE  M 
PE  M   .47  .37  .57
PE  M   0.27
Declaration and notation: 2pts
E =Student needs English Remediation.
M =Student needs Math Remediation.
5b.) What is the probability that a 2005 Freshman needs remediation in Math but not English?
Use probability notation to express your answer. 8pts
𝑃(𝑀 ∩ 𝐸̅ ) = 𝑃(𝑀) − 𝑃(𝑀 ∩ 𝐸)
𝑃(𝑀 ∩ 𝐸̅ ) = 0.37 − 0.27
𝑃(𝑀 ∩ 𝐸̅ ) = 0.10
5c.) Find the probability that either 6 or 7 of 10 CSU Freshman will not require remediation class work.
Declare the variable. Use probability notation to express your answer. 8pts
X = The number of 10 CSU freshman that will not require remediation. 1pt
10 
10 
P6  X  7    0.436 0.57 4   0.437 0.57 3
6 
7 
P6  X  7  0.140129 + 0.060407 = 0.200536
Stat 109
Sample Exam 1A KEY
541
5d.) Find the first standard deviation window to express the typical range of 10 randomly chosen
Freshman that will not require remediation?
8pts
np  np1  p  = 10  0.43  10  0.431  0.43  4.3  1.56556  2.734, 5.866 Notation 1pt
“About 2.7 to 5.9 of 10 Freshman will typically not require remediation.”
6.) Given that one out of four salamanders are consumed as prey before reaching sexual maturity, Find the
probability that between 10% and 30% of 50 salamanders will be consumed before reaching maturity. Show
all work and use proper notation for full credit. Finish with an English sentence.
14pts
Y
 pˆ  pˆ  0.5
Method 1: Declare proportions with the continuity correction.
n
n
Declare the parameter: 𝑝̂ = proportion of 50 drawn salamanders that are juveniles. 2pt
x = Salamanders
x = consumed salamanders
x = Proportion of salamanders
that are consumed.
P = portion consumed.
-2pt for all of these inadequate declarations.
If you are going to use calculations that determine the probability
that a sample proportion will range between some specified
̂, that we must declare. Do
thresholds, then this is this variable, 𝒑
not use “x” or “y” unless you are declaring for counts and plan to
calculate accordingly (see the next page.) Be sure to include the
sample size as this will affect probability.

P0.1  pˆ  0.3  P 0.10 
0.5
0.5 
 pˆ  0.30 

50
50 

P0.1  pˆ  0.3  P0.09  pˆ  0.31
Convert proportions to Z scores.
Use:
P0.1  pˆ  0.3  P????  Z  ???


0.09  .25
Z
P0.1  pˆ  0.3  P
0.25  1  0.25

50

P0.1  pˆ  0.3  P 2.613  Z  0.9798
P0.1  pˆ  0.3  P( Z  0.98)  PZ  2.61
P0.1  pˆ  0.3  0.836457 – 0.004527
P0.1  pˆ  0.3  0.83192
Z
pˆ  p
p(1  p)
n


0.31  .25

0.25  1  0.25 

50

Answer in English: “There is about
an 83% chance that between 10% and
30% of the 50 salamanders will be
consumed before reaching maturity.
Stat 109
Sample Exam 1A KEY
542
Problem #6) Method 2: Convert proportions to counts, then use the continuity correction.
̂=
𝒑
𝒀
→
𝒏
̂∙𝒏
𝒀=𝒑
Declare the parameter: Y = number of 50 drawn salamanders that are consumed. 2pt
x = Salamanders
x = consumed salamanders
x = Proportion of salamanders
that are consumed.
P = juveniles.
-2pt for all of these inadequate declarations.
If you are going to use calculations that determine the probability
that a sample count will range between some specified thresholds,
then this is the variable that we must declare. Be sure to include
the sample size as this will affect probability.
𝑃(0.1 ≤ 𝑝̂ ≤ 0.3) → 𝑃(0.1 ∙ 50 ≤ 𝑝̂ ∙ 𝑛 ≤ 0.3 ∙ 50) → 𝑃(5 ≤ 𝑌 ≤ 15)
𝑃(5 ≤ 𝑌 ≤ 15) ≈ 𝑃(5 − 0.5 ≤ 𝑌 ≤ 15 + 0.5) = 𝑃(4.5 ≤ 𝑌 ≤ 15.5)
After applying the continuity correction we convert the counts of juvenile
salamanders to Z-scores using the binomial expressions for the mean and
standard deviation: 𝜇 = 𝑛 ∙ 𝑝 and 𝜎 = √𝑛 ∙ 𝑝(1 − 𝑝). Note that “one out of
four” implies that p = 0.25
𝑃(4.5 ≤ 𝑌 ≤ 15.5) = 𝑃 (
4.5 − 0.25(50)
√50(0.25)(1 − 0.25)
≤𝑍≤
𝑍=
𝑍=
𝑋−𝜇
𝜎
𝑋−𝑛∙𝑝
√𝑛 ∙ 𝑝(1 − 𝑝)
15.5 − 0.25(50)
√50(0.25)(1 − 0.25)
)
𝑃(4.5 ≤ 𝑌 ≤ 15.5) = 𝑃(−2.61279 ≤ 𝑍 ≤ 0.979796) ≈ 𝑃(−2.61 ≤ 𝑍 ≤ 0.98)
We round to the nearest Hundredths for the Z-table probability values.
𝑃(4.5 ≤ 𝑌 ≤ 15.5)  P 2.613  Z  0.9798
𝑃(4.5 ≤ 𝑌 ≤ 15.5)  P( Z  0.98)  PZ  2.61
𝑃(4.5 ≤ 𝑌 ≤ 15.5) ≈ 0.836457 – 0.004527
𝑃(4.5 ≤ 𝑌 ≤ 15.5) ≈ 0.83192
Answer in English: “There is about an 83% chance that between 10% and 30% of the 50
salamanders will be consumed before reaching maturity.
Stat 109
Sample Exam 1B
Name__________
543
1.) Phenotypic risk for type 2 diabetes was assigned to 2377 people of European ancestry in the Framingham
Offspring Study. N Engl J Med. Nov 20, 2008; 359(21): 2208–2219. Participants were genotyped for 18 SNPs
(single nucleotide polymorphisms) associated with type 2 diabetes. Based upon the frequency of these 18 SNPs
a phenotypic risk for type 2 diabetes was given to each participant on a scale that ran from a low of 7 to a high
of 27. Three categories of phenotypic risk for type 2 diabetes were based upon one’s phenotypic score: Low at
less than 15, Medium for scores between 16 and 20 inclusive, and High for scores greater than 20. Europeans
over 50 years of age with a low phenotypic score had a 7% incidence of type 2 diabetes. While Europeans over
50 with a high phenotypic score had a 17% incidence of type 2 diabetes. Given that a sub-population of folks
with European ancestry over 50 years of age contains people with only high and low phenotypic scores, 30% of
which were high, with the rest having low scores, and that environmental factors such as diet and exercise have
been controlled, determine the following using correct probability notation.
Declare variables (2pts)
a.)
Find the probability that an over 50 year old with European ancestry will have Type 2 diabetes. 7pts
b)
Find the probability that an over 50 year old with European ancestry will have a low phenotypic
score given that the person has Type 2 diabetes. 7pts
c) Is the incidence of type 2 diabetes independent of the phenotypic score for people of European
ancestry? (Again use probability notation and numerical values to support your answer.) 5pts
Stat 109
Sample Exam 1B
2.) Instrumental conditioning is a term developed by Edward Thorndike (1898) in
which an animal gradually learns a task through repetition. A Thorndike box was
first used on house cats. The cat must learn how to paw at a series of levers and
latches with several false starts before she can escape from the box. Given that a
large number of individual trials were observed for cats placed in a Thorndike box,
suppose that each cat was repeatedly subjected to a Thorndike box until her escape
from the box could be performed in under a minute. A pdf of the number of trials
required for each cat to reach an escape time of less than a minute was recorded and
is posted above.
544
x, trials
4
5
6
7
8
9
10
p(x)
0.1
0.2
0.3
0.0
0.2
0.0
0.2
2a.) Find the expected number of trials required before a house cat can escape from the box in
less than a minute.(the mean of x.)
5 pts
2b.) Find the standard deviation of the number of trials required before a house cat can escape
from the box in less than a minute. 8pts
2c.) Find the typical range of the number of trials a cat needs to escape the Thorndike box in less
than a minute as the first standard deviation of x. (show your work.) 4pts (1 pt for notation)
Stat 109
Sample Exam 1B
3.) Kiama Blowhole Eruption Intervals:
An ocean swell produces spectacular eruptions of water through a hole in
the cliff at Kiama, about 120km south of Sydney, Australia, known as the
Blowhole. The times in seconds between each of 64 successive eruptions
from 1340 hours on 12 July 1998 were observed using a digital watch.
3a.) Circle the distribution that best describes the interval data for the
blowhole eruptions:
Skewed Left
Symmetrical
545
7
10
15
18
28
40
60
83
8
10
16
21
29
42
61
83
8
10
17
25
29
47
61
87
8
10
17
25
34
51
68
89
8
11
17
26
35
54
69
91
8
11
18
27
36
55
73
95
9
12
18
27
36
56
77
146
9
14
18
28
37
60
82
169
Skewed Right
3b.) Use correct notation to express the
5 key values for a boxplot for the
interval of time in seconds between
eruptions at the Kiama Blowhole.
11 pts
3c.) Draw a boxplot of the Kaima Blowhole data. Properly denote any outliers in the plot.
Stat 109
Sample Exam 1B
546
4.)
A population of 15 ruby throated hummingbirds, Archilochus
Declare the event variable.
colubris, was observed in a controlled environment as the subject of a student
senior’s thesis. Of interest was the quantitative feeding behavior of humming
birds when a food source is plentiful at numerous sites. Eight humming bird
feeders were filled with a sugar-nectar solution suspended from scales that
would digitally record the mass of each humming bird feeder after each
successive feeding. The difference in the feeder mass between each successive
recording provided a record of the amount of fluid consumed by a humming
bird at each feeding. The data of the mass consumed at each feeding was
normally distributed with a mean of 0.25 g and a standard deviation of 0.06 g.
Answer the following questions using the event variable with probability
notation and show the equivalent Z score within probability notation.
4a.)
4a.) At what percentile is a feeding of 0.2 grams? 8pts
4b.) The upper tenth percentile of feeding mass corresponds to what mass? 8pts
4c.) How much mass is consumed if the feeding
occurred at the 72nd percentile of feeding masses? 8pts
Stat 109
Sample Exam 1B
4d.) The same senior thesis on the behavior of feeding for the ruby
throated humming bird tracked the number of feedings made by each
of the 15 hummingbirds in an hour. The feedings taken in an hour by
each bird was normally distributed with a mean of 11.2 and a standard
deviation of 1.8. Determine the following probabilities using correct
notation.
547
Declare the event variable. 1pt
4e.) Find the probability that a randomly drawn hummingbird makes more than12 feedings in an hour. 8pt
4f.)
Find the probability that a randomly drawn humming bird feeds less than 10 times in an hour. 8pt
Stat 109
Sample Exam 1B
4g.) Find the probability that a randomly drawn humming bird feeds anywhere between 9 and 13 times
(inclusive) in a given hour. 8pt
4h.) Given that a randomly drawn humming bird feeds at the upper 4th percentile of feeding frequency,
how many times does the feed in an hour on average? 8pt
548
Stat 109
Sample Exam 1B
549
5a.)
The Colorado pike minnow (AKA white or Colorado river salmon, Ptychocheilus lucius) is the
largest minnow native to North America, and it is well known for its spectacular fresh water spawning
migrations and homing ability. Despite a massive recovery effort, its numbers decline. Hampered by a loss of
habitat, the young of this once abundant fish is overwhelmed in its nursery habitat by invasive small fishes
(such as red shiner and fathead minnow). Sites sampled from over 75 tributaries of the Colorado river found
that both the invasive species of red shiner or fathead minnow was present in 38% of sampled sites. Given that
55% of the river sites had red shiners present and that 47% of the sampled sites had fathead minnows present,
what portion of the sampled sites had either invasive species present? Use probability notation to express
your answer. 6pts
5b.) What is the probability that a tributary site of the Colorado has fathead minnows but not red shiners?
Use probability notation to express your answer. 6pts
5c.) What is the probability that a tributary site of the Colorado has either both invasive species of fathead
minnows and red shiners present or has neither invasive species present?
Use probability notation to express your answer. 8pts
Stat 109
Sample Exam 1B
550
5d.) Find the probability that more than 6 of 9 Colorado tributaries have either red shiner or fat head
minnow presence. Declare the variable. Use probability notation to express your answer. 8pts
5e.)
Find the probability that at least 8 of 9 Colorado tributaries have fat head minnow presence. 8pts
5e.)
Find the probability that a majority of 9 Colorado tributaries have red shiner presence. 8pts
5f.)
Find the first standard deviation window to express the typical range of 9 Colorado tributaries that
have red shiner presence.
8pts
𝑎. ) 𝑥̅ 𝑎𝑛𝑑 𝜇 ?
5g.) Describe the difference in meaning between these symbols:
𝑏. ) 𝑝̂ 𝑎𝑛𝑑 𝑝 ?
𝑐. ) 𝑥̃ 𝑎𝑛𝑑 𝜂 ?
Stat 109
Sample Exam 1B
551
6.) The 4 basic categories of human blood types (O, A, B, and AB) are coupled with an Rh factor that is
denoted with a plus (+) for its presence or minus sign (−) for its absence. This Rh factor is found
predominantly in rhesus monkeys, and to varying degree in human populations. For the US population it is
present in 83.3% of the population. Given that 40 people from the United States are randomly drawn, what
is the probability the sample proportion has between 80% to 90% of folks with an Rh + factor for their
blood type? Show all work and use proper notation for full credit. Finish with an English sentence.
14pts
Stat 109
Sample Exam 1B
Solution
552
1.) Phenotypic risk for type 2 diabetes was assigned to 2377 people of European ancestry in the Framingham
Offspring Study. N Engl J Med. Nov 20, 2008; 359(21): 2208–2219. Participants were genotyped for 18 SNPs
(single nucleotide polymorphisms) associated with type 2 diabetes. Based upon the frequency of these 18 SNPs
a phenotypic risk for type 2 diabetes was given to each participant on a scale that ran from a low of 7 to a high
of 27. Three categories of phenotypic risk for type 2 diabetes were based upon one’s phenotypic score: Low at
less than 15, Medium for scores between 16 and 20 inclusive, and High for scores greater than 20. Europeans
over 50 years of age with a low phenotypic score had a 7% incidence of type 2 diabetes. While Europeans over
50 with a high phenotypic score had a 17% incidence of type 2 diabetes. Given that a sub-population of folks
with European ancestry over 50 years of age contains people with only high and low phenotypic scores, 30% of
which were high, with the rest having low scores, and that environmental factors such as diet and exercise have
been controlled, determine the following using correct probability notation.
Declare variables (2pts)
D = Event that a European has Type 2 diabetes.
L = Event that a European has a low SNP score.
H = Event that a European has a high SNP score.
a.)
Find the probability that an over 50 year old with European ancestry will have Type 2 diabetes.
PD   PD L PL  PD H  PH 
PD  0.07  0.7  0.17  0.3
PD  .10
c)
Notation: 2pts
Calculation: 5pts
“About 10% of Europeans over 50 will have Type 2 diabetes.”
Find the probability that an over 50 year old with European ancestry will have a low phenotypic
score given that the person has Type 2 diabetes. 7pts
P L D  
P L D  
P D L  P L 
P L  D 

P D 
P D 
0.07  0.70
= 0.49
0.10
Notation: 2pts
Calculation: 5pts
“About 49% of over 50 year olds with European ancestry with Type 2 diabetes will have a
low phenotypic score.”
c) Is the incidence of type 2 diabetes independent of the phenotypic score for people of European
ancestry? (Again use probability notation and numerical values to support your answer.) 5pts
Stat 109
Sample Exam 1B
Solution
553
1c.) Continued..
For independent events the following must be true: P A  B  P A PB
We will use the Type 2 diabetes and low phenotypic scores since it is at hand:
PL  D   PL   PD  ? Is this true?
???
PL  D   PD L PL from above
PL  D  0.07  0.7
PL  D  0.049
Since 0.049  0.07
Then: PL  D  PL  PD
Therefore we do not have independent
events.
PD  PL  0.1 0.70
PD  PL  0.07
In English:
“The probability of whether an over 50 year old
of European ancestry develops Type 2 diabetes
depends on their phenotypic score.”
2.) Instrumental conditioning is a term developed by Edward Thorndike (1898)
in which an animal gradually learns a task through repetition. A Thorndike box
was first used on house cats. The cat must learn how to paw at a series of levers
and latches with several false starts before she can escape from the box. Given
that a large number of individual trials were observed for cats placed in a
Thorndike box, suppose that each cat was repeatedly subjected to a Thorndike box
until her escape from the box could be performed in under a minute. A pdf of the
number of trials required for each cat to reach an escape time of less than a minute
was recorded and is posted above.
x, trials
4
5
6
7
8
9
10
p(x)
0.1
0.2
0.3
0.0
0.2
0.0
0.2
2a.) Find the expected number of trials required before a house cat can escape from the box in
less than a minute.(the mean of x.)
5 pts
   xi pxi   4  0.1  5  0.2  6  0.3  7  0.0  8  0.2  9  0.0  10  0.2 = 6.8 
1
pt for Notation
2
  6.8
2b.) Find the standard deviation of the number of trials required before a house cat can escape
from the box in less than a minute. 8pts

 x px   
2
i
2
i
 4 2 .1  52 .2   6 2 .3  7 2 0  82 .2   9 2 0  10 2 .2   6.82  3.96  1.99

  1.99
Stat 109
2c.)
Sample Exam 1B
Solution
554
Find the typical range of the number of trials a cat needs to escape the Thorndike box in less than
a minute as the first standard deviation of x. (show your work.) 4pts (1 pt for notation)
𝝁 ± 𝝈 = 𝟔. 𝟖 ± 𝟏. 𝟗𝟗 = (𝟒. 𝟖𝟏, 𝟖. 𝟕𝟗)
3.) Kiama Blowhole Eruption Intervals:
An ocean swell produces spectacular eruptions of water through a hole in
the cliff at Kiama, about 120km south of Sydney, Australia, known as the
Blowhole. The times in seconds between each of 64 successive eruptions
from 1340 hours on 12 July 1998 were observed using a digital watch.
3a.) Circle the distribution that best describes the interval data for the
blowhole eruptions:
Skewed Left
Symmetrical
7
10
15
18
28
40
60
83
8
10
16
21
29
42
61
83
8
10
17
25
29
47
61
87
8
10
17
25
34
51
68
89
8
11
17
26
35
54
69
91
8
11
18
27
36
55
73
95
9
12
18
27
36
56
77
146
9
14
18
28
37
60
82
169
Skewed Right
3b.) Use correct notation to express the
5 key values for a boxplot for the
interval of time in seconds between
eruptions at the Kiama Blowhole.
11 pts
3b1.) Find the median:
𝑛 + 1 𝑡ℎ
𝑥̃ = (
)
2
3b2.) Determine the Quartile Criterion
𝐼𝑓 𝑁 ÷ 4, 𝑊𝑒 𝐴𝑣𝑒𝑟𝑎𝑔𝑒:
𝑡ℎ
𝑛 𝑡ℎ
𝑛
[4] +[4+1]
𝑄1
64 + 1 𝑡ℎ
=(
)
2
65 𝑡ℎ
=(
)
2
𝐼𝑓 𝑁 𝑖𝑠 𝑛𝑜𝑡 ÷ 4
[
2
𝑡ℎ
3𝑛 𝑡ℎ
3𝑛
[ 4 ] +[ 4 +1]
2
𝑄3
𝑡ℎ
𝑛
+1]
4
𝑡ℎ
3𝑛
[
+1]
4
= 32.5𝑡ℎ
= 32𝑛𝑑 + 0.5(33𝑟𝑑 − 32𝑛𝑑 )

= 28 + 0.5(28 − 28)

= 28 + 0.5(0)
𝑥̃ = 28

Where square brackets indicate that we round any
decimal down to find the nth value in the data set.
N = 64 is divisible by 4 so:
𝑄1 =
[
𝑡ℎ
𝑛 𝑡ℎ
𝑛
] +[ +1 ]
4
4
2
𝑎𝑛𝑑 𝑄3 =
[
𝑡ℎ
3𝑛 𝑡ℎ
3𝑛
] +[ +1 ]
4
4
2
Stat 109
Sample Exam 1B
Solution
555
3b3.) Find the 1st and 3rd Quartiles.
𝑡ℎ
𝑛 𝑡ℎ
𝑛
[4] +[4+1]
𝑄1 =
2
𝑡ℎ
64 𝑡ℎ
64
[
] +[ 4 +1]
𝑄1 = 4
2
3.) Find the 1st and 3rd Quartiles, Continued:
𝑡ℎ
3𝑛 𝑡ℎ
3𝑛
[ 4 ] +[ 4 +1]
𝑄3 =
2
𝑡ℎ
3 ∙ 64 𝑡ℎ
3 ∙ 64
]
+
[
+
1
]
4
4
𝑄3 =
2
[
3b4.) Find the IQR and Step.
[ 16 ]𝑡ℎ + [ 17 ]𝑡ℎ
2
14 + 15
𝑄1 =
2
𝑄1 =
𝑄1 = 14.5
[ 48 ]𝑡ℎ + [ 49 ]𝑡ℎ
2
60 + 60
𝑄3 =
2
𝑄3 =
𝑄3 = 60
3b5.) Find the Outlier Thresholds
IQR = Inter-quartile range
LOT = 𝑄1 − 𝑆𝑡𝑒𝑝
IQR = Q3 – Q1
LOT = 14.5 − 68.25
IQR = 60 – 14.5
LOT = −53.75
IQR = 45.5
UOT = 𝑄3 + 𝑆𝑡𝑒𝑝
𝑆𝑡𝑒𝑝 = 1.5 × 𝐼𝑄𝑅
UOT = 60 + 68.25
𝑆𝑡𝑒𝑝 = 1.5 × 45.5
𝑆𝑡𝑒𝑝 = 68.25
UOT = 128.25
3c.) Draw a boxplot of the Kaima Blowhole data. Properly denote any outliers in the plot.
Note! All data points that lie outside the outlier thresholds are expressed with
asterisks. The boxplot whiskers terminate at the last data point to lie within the
outlier thresholds. It is an error to extend the whiskers to the outlier thresholds.
*
*
Stat 109
Sample Exam 1B
Solution
4.)
A population of 15 ruby throated hummingbirds, Archilochus
colubris, was observed in a controlled environment as the subject of a
student senior’s thesis. Of interest was the quantitative feeding behavior of
humming birds when a food source is plentiful at numerous sites. Eight
humming bird feeders were filled with a sugar solution suspended from
scales that would digitally record the mass of each humming bird feeder
after each successive feeding. The difference in the feeder mass between
each successive recording provided a record of the amount of fluid
consumed by a humming bird at each feeding. The data of the mass
consumed at each feeding was normally distributed with a mean of 0.25 g
and a standard deviation of 0.06 g.
4a.) Answer the following questions using the event variable with probability
notation and show the equivalent Z score within probability notation.
4a.) At what percentile is a feeding of 0.2 grams? 8pts
X 
0.2  0.25
 Z 
Z = -0.8333, then
Z
0.6

556
Declare the event variable.
X = grams of sugar consumed
by a humming bird at a single
feeding. 1pt
For a vague declaration
without units of measure:
X = food. -1pt
X = bird. -1pt
X = sugar -1pt
Note! Here is a number with a
circle drawn around it:
0.203269
-1pt
This number has no meaning
without its supporting notation.
P X  0.2  PZ  0.83  0.203269
Answer like this with complete notation, Not like this:
Note that even if the prompt was in percentiles, we answer with probability notation.
4b.) The upper tenth percentile of feeding mass corresponds to what mass? 8pts
Note: This is an upper tail probability and we must know
to use the compliment value because the normal distribution
tables only provide for lower tail probabilities.
P X  ???   PZ  ???   0.10
a) Search the body of the table for Z score associated with 0.90.
b) Note that 0.899727 is closer to 0.90 than 0.901475. If you do not see this try subtracting 0.90 from both
values. Then we take the associated Z-score of 1.28 for 0.899727 as our closest estimate for 0.90.
c) Then PZ  1.28  0.10
d) Find the X associated with this Z score: Z 
X 

 1.28 
X  0.25
 X = 0.3268
0.06
e) Report the answer using probability notation.
Show the transition from the real world X values to the associated Z scores.
P X  0.3268  PZ  1.28  0.10
Stat 109
Sample Exam 1B
Solution
557
4c.) How much mass is consumed if the feeding
occurred at the 72nd percentile of feeding masses? 8pts
Note that the percentiles on the
normal curve run from zero to
100% and accumulate from left to
right.
The corresponding masses for these
percentiles of feedings are shown below.
a) Search the body of the table for Z
score associated with 0.72.
c) We find:
d) Using Standardization we find x:
x
x  0.25
PZ  0.58  0.72
Z
 0.58 

0.6
e) Report your answer with full probability notation:
P X  0.2848  PZ  0.58  0.72
b) P X  ???   PZ  ???   0.72
4d.) The same senior thesis on the behavior of feeding for the ruby
throated humming bird tracked the number of feedings made by each
of the 15 hummingbirds in an hour. The feedings taken in an hour by
each bird was normally distributed with a mean of 11.2 and a standard
deviation of 1.8. Determine the following probabilities using correct
notation.
b) Declare the event variable. 1pt
N = The number of feedings
made by a humming bird
in an hour.
N = feedings -1pt:
For a vague declaration
4e.) Find the probability that a randomly drawn hummingbird makes more than12 feedings in an hour. 8pt
PN  12  PN  13
Note that discrete counts must be adjusted for a mapping to the continuous
normal curve. Strict inequalities of “more than” must be converted
numerically to “more than or equal to” if using a continuous distribution.
PN  12  PN  13  PN  12.5
Z
x


12.5  11.2
1.8
Z  0.72
PZ  0.72  0.764237
PN  12  1  PZ  0.72  0.235763
A continuity correction is necessary here because a
discrete count of visits to a feeder must be mapped to a
continuous curve of the normal distribution. Note that we
always choose a half count in the direction that enlarges
the shaded region beneath the normal curve. Also note
that a compliment of one minus the Z-table probability
must be accounted for here.
In English: “About 23.6% of the time a
randomly drawn humming bird will make more
than 12 feedings in an hour.
Answer with complete notation like this, and not this:
0.235763
Stat 109
4f.)
Sample Exam 1B
Solution
558
Find the probability that a randomly drawn humming bird feeds less than 10 times in an hour. 8pt
PN  10  PN  9
Note that discrete counts must be adjusted for a mapping to the continuous
normal curve. Strict inequalities of “less than” must be converted
numerically to “less than or equal to” if using a continuous distribution.
PN  10  PN  9  PN  9.5
Z
x


A continuity correction is necessary here because a
discrete count of feedings must be mapped to a
continuous curve of the normal distribution. Note that we
always choose a half count in the direction that enlarges
the shaded region beneath the normal curve.
9.5  11.2
1.8
Z  0.9444
PZ  0.94  0.173609
In English: “About 17.4% of the time a
randomly drawn humming bird will feed less
than 10 times in an hour.”
PN  10  PZ  0.94  0.173609
Answer with complete notation like this, and not this:
0.173609
4g.) Find the probability that a randomly drawn humming bird feeds anywhere between 9 and 13 times
(inclusive) in a given hour. 8pt
P9  N  13
Note that discrete counts must be adjusted for a mapping to the continuous normal curve.
Here the inclusive counts (of 9 and 13) are already specified in the problem.
P9  N  13  P8.5  N  13.5
A continuity correction is still necessary because a
discrete count of visits to the feeder must be mapped to
a continuous curve of the normal distribution. Note
that we always choose a half count in the direction that
enlarges the shaded region beneath the normal curve.
Z Low 
x

Z Low  1.5
Continued

8.5  11.2
1.8
Z High 
x


Z High  1.2778
13.5  11.2
1 .8
Stat 109
Sample Exam 1B
Solution
559
4g.) Find the probability that a randomly drawn humming bird feeds anywhere between 9 and 13 times
(inclusive) in a given hour. 8pt Continued
P9  N  13  P 1.50  Z  1.28
In English: “About 83.3% of the time a
randomly drawn humming bird will feed
between 9 and 13 times in an hour.
P9  N  13  PZ  1.28  PZ  1.50
P9  N  13  0.899727  0.066807
Answer with complete notation
like this, and not this
P13  N  19  0.83292
P9  N  13  P 1.50  Z  1.28  0.83292
0.83292
4h.) Given that a randomly drawn humming bird feeds at the upper 4th percentile of feeding frequency,
how many times does the feed in an hour on average? 8pt
PN  ??  PZ  ??  0.04
The prompt for an upper 4th percentile
means that we must search the body of
the table for the “lower” 96th percentile as
the Z-tables will only give lower tailed
probabilities.
a) Search the body of the table for Z
score associated with 0.96.
b) P X  ???   PZ  ???   0.96
c) We find:
PZ  1.75  0.96
e) Solve for x and report your answer
with full probability notation:
f)
PN  14.85  PZ  1.75  0.04
d) Using Standardization we find x. But careful here! We
must provide for the continuity correction and for an
upper tail this means that we must subtract 0.5 from the
lower threshold to expand the probability space by half a
visit to a humming bird feeder.
x
x  0.5  11.2 x  11.7
Z
 1.75 


1.8
1.8
x  14.85
A humming bird at the upper 4th percentile will feed an
average of 14.85 times per hour.
Stat 109
Sample Exam 1B
Solution
560
5a.)
The Colorado pikeminnow (AKA white or Colorado river salmon, Ptychocheilus lucius) is the
largest minnow native to North America, and it is well known for its spectacular fresh water spawning
migrations and homing ability. Despite a massive recovery effort, its numbers decline. Hampered by a loss of
habitat, the young of this once abundant fish is overwhelmed in its nursery habitat by invasive small fishes
(such as red shiner and fathead minnow). Sites sampled from over 75 tributaries of the Colorado river found
that both the invasive species of red shiner or fathead minnow was present in 38% of sampled sites. Given that
55% of the river sites had red shiners present and that 47% of the sampled sites had fathead minnows present,
what portion of the sampled sites had either invasive species present? Use probability notation to express
your answer. 6pts
PR  F   PR  PF   PR  F 
PR  F   0.55  0.47  0.38
PR  F   0.64
Declaration and notation: 2pts
R =Red shiners are present in sample.
F =Fatheads are present in sample.
5b.) What is the probability that a tributary site of the Colorado has fathead minnows but not red shiners?
Use probability notation to express your answer. 6pts
𝑃(𝐹 ∩ 𝑅̅ ) = 𝑃(𝐹) − 𝑃(𝐹 ∩ 𝑅)
𝑃(𝐹 ∩ 𝑅̅ ) = 0.47 − 0.38
𝑃(𝐹 ∩ 𝑅̅ ) = 0.09
5c.) What is the probability that a tributary site of the Colorado has either both invasive species of fathead
minnows and red shiners present or has neither invasive species present?
Use probability notation to express your answer. 8pts
𝑃((𝐹 ∩ 𝑅) ∪ (𝐹̅ ∩ 𝑅̅ )) = 1 − 𝑃(𝐹) − 𝑃(𝑅) + 2𝑃(𝐹 ∩ 𝑅)
𝑃((𝐹 ∩ 𝑅) ∪ (𝐹̅ ∩ 𝑅̅ )) = 1 − 0.47 − 0.55 + 2 ∙ 0.38
𝑃((𝐹 ∩ 𝑅) ∪ (𝐹̅ ∩ 𝑅̅ )) = 0.74
5d.) Find the probability that more than 6 of 9 Colorado tributaries have either red shiner or fat head
minnow presence.
Declare the variable. Use probability notation to express your answer. 8pts
X = The number of 9 Colorado tributary sites that have either invasive species. 1pt
 9
 9
 9
P X  6  P X  7   0.64 7 0.362   0.6480.361   0.649 0.360
 7
8 
 9
P X  6  0.205195 + 0.091198 + 0.018014 = 0.31441
Stat 109
5e.)
Sample Exam 1B
Solution
561
Find the probability that at least 8 of 9 Colorado tributaries have fat head minnow presence. 8pts
X = The number of 9 Colorado tributary sites that have fat head minnows. 1pt
 9
 9
P X  8   0.4780.531   0.479 0.530
8 
 9
P X  8  0.011358 + 0.001119 = 0.012477
5e.)
Find the probability that a majority of 9 Colorado tributaries have red shiner presence. 8pts
X = The number of 9 Colorado tributary sites that have red shiners. 1pt
 9
 9
 9
 9
9
P X  5   0.5550.454   0.556 0.453   0.557 0.452   0.5580.451   0.559 0.450
5
 6
 7
8 
9
P X  5  0.260036 + 0.211881 + 0.110986 + 0.033912 + 0.004605 = 0.62142
5f.)
Find the first standard deviation window to express the typical range of 9 Colorado tributaries that
have red shiner presence.
8pts
= 9  0.55  9  0.551  0.55  4.95  1.492481  3.458, 6.172 Notation 1pt
“About 3.5 to 6.2 of 9 Colorado tributaries will have red shiner presence.”
𝑎. ) 𝑥̅ 𝑎𝑛𝑑 𝜇 ?
5g.) Describe the difference in meaning between these symbols:
𝑏. ) 𝑝̂ 𝑎𝑛𝑑 𝑝 ?
𝑐. ) 𝑥̃ 𝑎𝑛𝑑 𝜂 ?
𝑎. )
𝑥̅ denotes the sample mean while 𝜇 denotes the population mean (or true mean).
𝑏. )
𝑝̂ denotes the sample proportion while p denotes the population proportion (or true proportion).
𝑐. )
𝑥̃ denotes the sample median while 𝜂 denotes the population median (or true median).
Stat 109
Sample Exam 1B
Solution
562
6.) The 4 basic categories of human blood types (O, A, B, and AB) are coupled with an Rh factor that is
denoted with a plus (+) for its presence or minus sign (−) for its absence. This Rh factor is found
predominantly in rhesus monkeys, and to varying degree in human populations. For the US population it is
present in 83.3% of the population. Given that 40 people from the United States are randomly drawn, what
is the probability the sample proportion has between 80% to 90% of folks with an Rh + factor for their
blood type? Show all work and use proper notation for full credit. Finish with an English sentence.
14pts
Y
 pˆ  pˆ  0.5
Method 1: Declare proportions with the continuity correction.
n
n
Declare the parameter:
𝑝̂ = proportion of 40 randomly drawn Americans with Rh+ blood types. 2pt
x = USA Rh+ blood
x = Rh+ blood types
x = Proportion of USA with Rh+
p = proportion of USA with Rh+
-2pt for all of these inadequate declarations.
If you are going to use calculations that determine the probability
that a sample proportion will range between some specified
̂, that we must declare. Do
thresholds, then this is this variable, 𝒑
not use “x” or “y” unless you are declaring for counts and plan
to calculate accordingly (see the next page.) Be sure to include
the sample size as this will affect probability.
0.5
0.5 

 pˆ  0.90 
P0.8  pˆ  0.9  P 0.80 

40
40 

P0.8  pˆ  0.9  P0.7875  pˆ  0.9125
Convert proportions to Z scores.
Use:
pˆ  p
P0.8  pˆ  0.9  P????  Z  ???


0.7875  .833
Z
P0.8  pˆ  0.9  P
 0.833  1  0.833

40

P0.8  pˆ  0.9  P 0.7715  Z  1.3481
P0.8  pˆ  0.9  P( Z  1.35)  PZ  0.77
P0.8  pˆ  0.9 = 0.911492 – 0.220650
P0.8  pˆ  0.9  0.690842
Z
p(1  p)
n


0.9125  .833 
0.833  1  0.833 

40

Answer in English: “There is about a
69.1% chance that between 80% and
90% of 40 randomly drawn Americans
will have Rh+ blood types.
Stat 109
Sample Exam 1B
Solution
563
Problem #6) Method 2: Convert proportions to counts, then use the continuity correction.
̂=
𝒑
𝒀
→
𝒏
̂∙𝒏
𝒀=𝒑
Declare the parameter: Y = number of 40 randomly drawn Americans with Rh+ blood types. 2pt
x = Rh+ blood
x = American Rh+ blood
x = Number of Americans
with Rh+ blood.
P = proportion of Rh+ blood.
-2pt for all of these inadequate declarations.
If you are going to use calculations that determine the probability
that a sample count will range between some specified thresholds,
then this is the variable that we must declare. Be sure to include
the sample size as this will affect probability.
𝑃(0.8 ≤ 𝑝̂ ≤ 0.9) = 𝑃(0.8 ∙ 40 ≤ 𝑝̂ ∙ 𝑛 ≤ 0.9 ∙ 40) = 𝑃(32 ≤ 𝑌 ≤ 36)
𝑃(32 ≤ 𝑌 ≤ 36) ≈ 𝑃(32 − 0.5 ≤ 𝑌 ≤ 36 + 0.5) = 𝑃(31.5 ≤ 𝑌 ≤ 36.5)
After applying the continuity correction we convert the counts of
Americans with Rh+ blood types to Z-scores using the binomial
expressions for the mean and standard deviation: 𝜇 = 𝑛 ∙ 𝑝 and 𝜎 =
√𝑛 ∙ 𝑝(1 − 𝑝). Note that p = 0.833.
𝑍=
𝑍=
𝑋−𝜇
𝜎
𝑋−𝑛∙𝑝
√𝑛 ∙ 𝑝(1 − 𝑝)
31.5 − 40(.833)
36.5 − 40(.833)
𝑃(31.5 ≤ 𝑌 ≤ 36.5) = 𝑃 (
≤𝑍≤
)
√40(0.833)(1 − 0.833)
√40(0.833)(1 − 0.833)
𝑃(31.5 ≤ 𝑌 ≤ 36.5) = 𝑃(−0.7715 ≤ 𝑍 ≤ 1.3481) ≈ 𝑃(−0.77 ≤ 𝑍 ≤ 1.35)
We round to the nearest Hundredths for the Z-table probability values.
Answer in English: “There is about a
P31.5  Y  36.5  P( Z  1.35)  PZ  0.77 69.1% chance that between 80% and
P31.5  Y  36.5 = 0.911492 – 0.220650
P31.5  Y  36.5  0.690842
90% of 40 randomly drawn Americans
will have Rh+ blood types.
Stat 109
Blank Sheet
Sample Exam 1B
Solution
564