Page 285
4. (a) P( X  2) = binompdf(5, 0.05, 2) = 0.201
(b) P( X  2)  1  P( X  2) = 1 − binomcdf(5, 0.05, 2) = 0.001
(c) P ( X  5)  binompdf(5, 0.05, 5) = 0.0000003
(d) Since the probability of each event gets less likely the probabilities decrease.
6. The probability of guessing correctly is
1
 0.2 . P( X  15)  1  P( X  14)
5
= 1 − binomcdf(20, 0.2, 14) = 0.0000002
This is very unlikely to happen.
8. Note that P(drops out) = 0.103
(a) P ( X  2) = binomcdf(10, .103, 2) = 0.925
(b) If at least six graduate then four or fewer drop out
P( X  4)  binomcdf(10, 0.103, 4) = 0.998
(c) P(none drop out) = binompdf(10, 0.103, 0) = 0.337
10. Note that P(both work) = 0.521
(a) P( X  0)  binompdf(5, 0.521, 0) = 0.025
(b) P( X  3)  1  P( X  3)  1 − binomcdf(5, 0.521, 3) = 0.215
(c) P( X  2)  P( X  1)  1 − binomcdf(5, 0.521, 1) = 0.162
12. Note that P(destination wedding) = 0.26
(a) P( X  6)  binompdf(12, 0.26,6) = 0.047
(b) P( X  6)  1  P( X  5) = 1 − binomcdf(12, 0.26, 5) = 0.06
(c) P( X  5)  P( X  4) = binomcdf(12, 0.26, 4) = 0.821
16.
  np  10  0.5  5
 2  npq  10  0.5  0.5  2.5
   2  2.5  1.581
18. Note that p  P(use e-mail)  0.83; q  1  p  0.17
  np  200  0.83  166
 2  npq  200  0.83  0.17  28.22
   2  28.22  5.31
24. Note that p  P(purchases online)  0.32; q  1  p  0.68
  np  200  0.32  64
 2  npq  200  0.32  0.68  43.52
Page 310
10. Area right of z  2.01  normalcdf(2.01, 10000)  0.0222
12. Area left of z  0.75  normalcdf(  10000,  0.75)  0.2266
16. Area between z  0.96 and z  0.36  normalcdf(  0.96,  0.36)  0.1909
18. Note when using normalcdf the left hand value must be entered first
Area between z  0.24 and z  1.12  normalcdf(  1.12, 0.24)  0.4634
28. P(1.23  z  0)  normalcdf(  1.23, 0)  0.3907
40. To find z use invnrom(area left of z).
Since the area left of 0 is 0.5 the area
z  invnorm(0.9066)  1.32
left of z = 0.5 + 0.4066 = 0.9066
Area = 0.48
z  invnorm(0.9761)  1.979
42. Area left of z = 1 − 0.0239 = 0.9761.
44. z  invnorm(0.9671)  1.840
46. (a) Area left of z = 0.5478. z  invnorm(0.5478)  0.12
(b) Area left of z = 0.6985. z  invnorm(0.6985)  0.52
(c) Area left of z = 0.8810. z  invnorm(0.8810)  1.18
48. The picture is:
Area = 0.48
Area = 0.26
Area = 0.26
−z
z
Area left of z = 0.74. z  invnorm(0.74)  0.643 . By symmetry the two values are z  0.643
Page 323
4. The 90th percentile has 90% of the area under the curve to the left of it. The z-score for the 90th
percentile is invnorm(0.9) = 1.282. Convert to X using x     z  1028  92(1.282)  1146
The z-score for 1200 =
1200  1028
 1.870 . Then P( X  1200)  P( z  1.870)
92
= normalcdf(1.870, 10000) = 0.0307
8. (a) The z-score for 15000 =
15000  12837
 1.442 . Then P( X  15000)  P( z  1.442)
1500
= normalcdf(1.442, 10000) = 0.747
13000  12837
 0.1087 .
1500
14000  12837
 0.7753 .
The z-score for 13000 =
1500
(b) The z-score for 13000 =
Then P(1300  X  14000)  P(0.1087  z  0.7753) = normalcdf(0.1087, 0.7753) = 0.2376
10. (a) The z-score for 30 =
30  25
 0.8197 . Then P( X  30)  P( z  0.8197)
6.1
= normalcdf(0.8197, 10000) = 0.206
(b) The z-score for 18 =
18  25
 1.1475 . Then P( X  18)  P( z  1.1475)
6.1
= normalcdf(−10000, −1.1475) = 0.1256
14. The z-score for 384 =
384  225
 3.5333 . Then P( X  384)  P( z  3.5333)
45
= normalcdf(3.5333, 10000) = 0.0002
18. The picture is:
Area = 0.5
Area = 0.25
Area = 0.25
z
−z
Area left of z = 0.75. z  invnorm(0.75)  0.6745 . By symmetry the two values are z  0.6745
The lower value is X     z  792  103(0.6745)  722.52
The upper value is X     z  792  103(0.6745)  861.47
Answer: $722.52 and $861.47
20. The picture is:
Area = 0.8
Area = 0.1
Area = 0.1
−z
z
Area left of z = 0.75. z  invnorm(0.9)  1.282 . By symmetry the two values are z  1.282
The lower value is X     z  246300  15000(1.282)  227070
The upper value is X     z  246300  15000(1.282)  265530
Answer: $227,070 to $265,530
26. The picture is:
For second part
For first part
Area = 0.3
Area = 0.1
z2
z1
Area left of z1 = 0.9. z  invnorm(0.9)  1.282 .
This one is sneaky: You are given the variance so take a square root to get the standard deviation
Then x     z  4.8  2.1(1.282)  6.658
Then 10% of hospital stays last longer than 6.658 days
Area left of z2 = 0.3. z  invnorm(0.3)  .5244 .
Then x     z  4.8  2.1(0.5244)  4.404
Then 30% of hospital stays last less than 4.404 days
30. The picture is:
Area = 0.2
z
Area left of z = 0.8. z  invnorm(0.8)  0.8416 .
Then x     z  64  9(0.8416)  71.6
Assuming scores are whole numbers the cut off score is 72
Page 330
x
45000  37764
 1.4188

5100
Then P( X  45000)  P( z  1.4188)  normalcdf(1.4188, 10000) = 0.0778
12. (a) z 

(b) For sample means the mean is the population mean  and the standard deviation is 
x 
38000  37764
 0.4007
5100
n
75
Then P( x  38000)  P( z  0.4007)  normalcdf(0.4007, 10000) = 0.3443
z


16. The standard deviation for samples of size 33 is 
x 
n
 2.6
33
23.8  24.3
 1.104

2.6
n
33
Then P( x  23.8)  P( z  1.1047)  normalcdf(10000, 1.1047) = 0.1346
z

20. The standard deviation for samples of size 34 is 
n
 4850
34
First part:
x 
50000  51803
 2.1677
4850
n
34
Then P( x  50000)  P( z  2.1677)  normalcdf(2.1677, 10000) = 0.9849
z


Second part:
x 
48000  51803
 4.5721

4850
n
34
Then P( x  48000)  P( z  4.5721)  normalcdf(10000, 4.5721) = .000002
z

22. (a) Use 120 for  and 5.6 for  . Then
P(120  x  121.8)
120  120
121.8  120
 P(
z
)
5.6
5.6
 P(0  z  .3214)
= normalcdf(0, .3214) = 0.126
(b) Use 120 for  and 5.6
30
for  . Then
n
P(120  x  121.8)
120  120
121.8  120
 P(
z
)
5.6
5.6
30
30
 P(0  z  1.7605)
= normalcdf(0, 1.7605) = .4608
(c) Means are less variable than individual data.
24. (a) Use 36.2 for  and 3.7 for  . Then
P(36  x  37.5)
36  36.2
37.5  36.2
z
)
3.7
3.7
 P(.0541  z  0.3514)
 P(
= normalcdf(0.0541, 0.3514) = 0.1589
(b) Use 36.2 for  and 3.7
for  . Then
15
P(36  x  37.5)
36  36.2
37.5  36.2
z
)
3.7
3.7
15
15
 P(.2094  z  1.3607)
 P(
= normalcdf(0.2094, 1.3607) = .4961
Page 364
10. From the TI83+ you find x  346.25 . For 92% confidence z  1.751 using invnorm(.96). [The
2
area to right of z is 0.04, so the area to the left is 0.96.] Then
2
x  z

   x  z

2
n
n
165.1
165.1
346.25  1.751(
)    346.25  1.751(
)
32
32
295.1    397.4
2
14. (a) A point estimate is what you get from one sample: 7.2 jobs
(b) For 95% confidence z  1.960 from the last line of the t-table or using invnorm(0.975)
2
x  z


   x  z
2
n
n
2.1
2.1
7.2  1.960(
)    7.2  1.960(
)
50
50
6.62    7.78
2
(c) For 99% confidence z  2.576 from the last line of the t-table or using invnorm(0.995)
2
x  z


   x  z
2
n
n
2.1
2.1
7.2  2.576(
)    7.2  2.576(
)
50
50
6.43    7.97
2
(d) To make it more likely that the interval contains the mean the interval must be wider.
18. For 90% confidence z  1.645 from the last line of the t-table or using invnorm(0.95)
2
x  z

   x  z
n
630
2
3987  1.645(

n
2
)    3987  1.645(
50
$3840    $4134
630
)
50
As a start up your pricing needs to be competitive so charge $3,800
20. For 95% confidence z  1.960 from the last line of the t-table or using invnorm(0.975)
2
x  z

2
190.7  1.960(
   x  z
n
54.2

2
n
)    190.7  1.960(
35
172.7    208.7
54.2
)
35
22. . For 90% confidence z  1.645 from the last line of the t-table or using invnorm(0.95)
2
x  z

2
58.0  1.645(
Page 372
n
4.8
   x  z

2
n
)    58.0  1.645(
171
57.40    58.60
4.8
)
171
6. From the TI83+ the sample standard deviation s  17.487 and the sample mean x  217.7
Since n =10 you have d.f. = n – 1 = 9. From the t-table t  2.262
2
s
s
   x  t
2
2
n
n
17.487
17.487
217.7  2.262(
)    217.7  2.262(
)
10
10
205.2    230.2
x  t
12. Since n =13 you have d.f. = n – 1 = 12. From the t-table t  3.055
2
s
s
x  t
   x  t
2
2
n
n
1.7
1.7
15  3.055(
)    15  3.055(
)
13
13
13.6    16.4
If you were to issue a warning to make sure people had enough time to get out of the way you
would use the highest speed of 16.4 mph.
14. Since n =8 you have d.f. = n – 1 = 72. From the t-table t  2.365
2
s
s
   x  t
2
2
n
n
4.1
4.1
13.1  2.365( )    13.1  2.365( )
8
8
9.7    16.5
x  t
16. Since n =8 you have d.f. = n – 1 = 72. From the t-table t  2.365
2
s
s
   x  t
2
2
n
n
4.1
4.1
13.1  2.365( )    13.1  2.365( )
8
8
9.7    16.5
x  t
20. Since n =24 you have d.f. = n – 1 = 23. From the t-table t  2.069
2
s
s
   x  t
2
2
n
n
7.5
7.5
46.1  2.069(
)    41.6  2.069(
)
24
24
38.4    44.8
x  t