Take Home Exam 2
Name -
Solutions
Please show all your work in the space provided.
1. What is the probability of each of the following? [12 pts]
a) Getting an 8 on roll of 2 dice.
5/36 = .139
b) Not getting a 7 on the roll of 2 dice.
1 - 6/36 = 1 - .167 = .833
c) Randomly selecting a spade or a face card from a deck of 52 cards.
P(A or B) = P(A) + P(B) – P(A and B) =
13/52 + 12/52 – 3/52 = 22/52 = 11/26 = .423
2a) Convert the distribution below to a probability distribution. [10 pts]
Passengers on transcontinental
Flight 385:
Passengers
Frequency
x
P(x)
159 – 193
5
159 – 193
5/30 =.167
194 – 228
9
194 – 228
9/30 = .3
229 – 263
6
229 – 263
6/30 = .2
264 – 298
4
264 – 298
4/30 = .133
299 – 333
6
299 – 333
6/30 = .2
b) What is the probability of there being at least 229 passengers on Flight 385?
.2 + .133
+ .2 = .533
_________
1
Take Home Exam 2
Name -
Solutions
3. Find the mean and standard deviation using the probability distribution from number
2 above. [12 pts]
x
P(x)
x·P(x)
(x – μ)
(x - µ)2
176
.167
29.39
-66.46
4416.93
737.63
211
.3
63.3
-31.46
989.73
296.92
246
.2
49.2
3.54
12.53
2.51
281
.133
37.37
38.54
1485.33
197.55
316
.2
63.2
73.54
5408.13
1081.63
(x - µ)2 · P(x)
μ = 242.46
x
176
211
246
281
316
P(x)
0.167
0.3
0.2
0.133
0.2
Σ
xP(x)
29.392
63.3
49.2
37.373
63.2
242.465
σ2 = 2316.23
x2P(x)
5172.992
13356.3
12103.2
10501.81
19971.2
61105.51
x2P(x)-μ2
2316.229
σ = 48.13
σ
48.12721
4. The questions below are based on the following statistics. [10 pts]
LaGuardia
Students
LGCC Students
who ride E Train
LGCC Students
who ride #7 train
Men
1090
450
540
Women
2150
875
615
a) What is the probability that a student selected at random is a man or a person who
rides the E train to school?
P(A or B) = P(A) + P(B) – P(A and B)
= (1090 / 3240) + (1325 / 3240) – (450 / 3240) = 1965/3240 = .60648
2
Take Home Exam 2
Name -
Solutions
b) What is the probability that a student selected at random is a woman who does not
ride the No. 7 train to school?
(2150 – 615) / 3240 = 1535/3240 = .47376
5. You are given that on average a particular insurance sales person is successful 15% of
the time. On a given week this person makes 12 sales calls. [20 pts]
a) What is the probability of making exactly 3 successful calls?
(12!)/(3!)(9!) = 12(11)(10) / 6 = 2(11)(10) = 220
220(.15)3(.85)9 = 220(.003375)(.2316) = .1735
b) What is the probability of making at least one successful call?
(12!)/(12!)(0!) = 1
P(0) = 1 (.15)0 (.85)12 = (.85)12 = .1422
P(at least 0) = 1 – P(0) = .8578
c) What is the mean and standard deviation given the information above?
μ = np
= 12(.15) = 1.8
σ = sqrt(npq) = sqrt(12(.15)(.85)) = sqrt(1.53) = 1.237
d) What are the maximum and minimum usual values?
μ + 2σ = 1.8 + 2(1.237) = 1.8 + 2.474 = 4.274
μ – 2σ = 1.8 – 2(1.237) = 1.8 – 2.474 = – 0.674
3
Take Home Exam 2
Name -
Solutions
6. Find the probability of z scores within the following ranges:[10 pts]
a)
P(– 1.23 < z < 1.23)
.8907 - .1093 = .7814
b) P(z < 2.1)
.9821
7. You are given that a score of 87 on an exam is at the 94th percentile and that the grades
are normally distributed. [6 pts]
a) What z score does this represent?
z = 1.555
b) If the mean score on the exam is 82, what is the standard deviation?
z = (x – u)/s = (87 – 82)/1.56 = 5/1.56 = 3.205
8. If the age of LaGuardia students is normally distributed with a mean of 24.3 and a
standard deviation is 3.6 years answer the following: [20 pts]
a) Convert each of the following ages to z scores.
29 years
z = (29 – 24.3) / 3.6 = 1.305
18 years
z = (18 – 24.3) / 3.6 = -1.75
b) What is the age of students with the following z scores?
– .85
x = sz + u = (3.6)(-.85) + 24.3 = 21.24
1.24
x = (3.6)(1.24) + 24.3 = 28.764
4
Take Home Exam 2
Name -
Solutions
c) What is the probability that the age of a given student is between 22 and 27 years of
age?
.7734 - .2611 = 0.5123
Bonus Question – 20 points!
Sample washer
The mean inside diameter of a population of 200 washers
produced by a machine is 0.502 inches and the standard
deviation is 0.005 inches. The company can only
accept washers whose inside diameters are in the range
from 0.496 in. to 0.508 in. otherwise the washers are
considered defective. Determine the percentage of
defective washers, assuming the diameters are normally
distributed.1
μ = .502 in.
23.01 %
µ = 0.502
σ = 0.005
x1 = 0.496
x2 = 0.508
z1 = (0.496 – 0.502) / .005 = – .006 / .005 = - 1.2
P ( z1 < -1.2) = .11507
z2 = (0.508 – 0.502) / .005 = .006 / .005 = 1.2
P ( z2 < 1.2) = .88493
P ( -1.2 < z < +1.2) = .88493 – .11507 = .76986
Percentage of Defective washers = 1 - .76986 = .23014 = 23.01%
1
a
5