Download Test 2 - La Sierra University

Math 251, 28 October 2003, Exam 2
Name:
ANSWERS
.
Instructions: Complete each of the following seven questions, and please explain and justify all
appropriate details in your solutions in order to obtain maximal credit for your answers.
1. At Kenwood College of Engineering, 45% of incoming freshmen students are female and 55% are
male. Recent records indicate that 80% of the entering female students will graduate with a BSE
degree, while 70% of the male students will obtain a BSE degree. If an incoming engineering student is
selected at random, find
(a) (1 pt)
(b) (2 pts)
(c) (1 pt)
(d) (2 pts)
(e) (2 pts)
(f) (2 pts)
Ans:
P(student will graduate, given student is female)
P(student will graduate, and student is female)
P(student will graduate, given student is male)
P(student will graduate, and student is male)
P(student will graduate)
P(student will graduate, or student is female)
Let F = event student is female, M=event student is male, G = event student will graduate, and
N = event student does not graduate.
(a)
(b)
(c)
(d)
(e)
P(G given F) = 0.80
P(F and G) = P(F)P(G given F) = 0.450.8 = 0.36
P(G given M) = 0.7
P(G and M) = P(M)P(G given M) = 0.550.70 = 0.385
P(G) = P(G given F) + P(G given M) (since a grad is either male or female)
= 0.36 + 0.385 = 0.745
(f) P(G or F) = P(G) + P(F) – P(G and F)
= 0.745 + 0.45 – 0.36 = 0.835
2. (a) (2 pts) President Geraty has recently received permission to excavate the site of an ancient
temple. In how many ways can he choose 8 of the 29 students in his History of Antiquities course to
join him on the dig?
Ans:
C29,8 =
29!
= 4,292,145
8!21!
(b) (2 pts) Of the 29 graduate students, 19 are female and 10 are male. In how many ways can President
Geraty select a group of 8 that consists only of females?
Ans:
C19,8 =
19!
= 75,582
8!11!
(c) (2 pts) What is the probability that President Geraty would randomly select a group of 8 consisting
of only females?
Ans:
The probability is the Answer in (b)  Answer in (a)  .0176
3. (a) (1 pt) Explain what mutually exclusive events are.
Ans:
They are events that cannot occur together, so the probability P(A and B) = 0 for mutually
exclusive events.
(b) (1 pt) Explain what independent events are.
Ans:
They are events such that the occurrence or nonoccurrence of one of the events does not affect
the probability of the other event occurring. In rules of probability, P(A given B) = P(B) and
P(B given A) = P(A) for independent events A and B.
(c) (2 pts) Give an example of two events that independent but are not mutually exclusive. Justify your
answer.
Ans:
A typical example is that of flipping coins. Let A=the event that the first toss is a head, and B =
the event that the second toss is a head. On a fair coin, A and B can occur together, in fact, P(A
and B) = 0.25. However, the outcome of A does not influence the outcome of B and vice versa,
in fact, P(A) = P(A given B) = 0.50 and P(B) = P(B given A) = 0.50.
4. (a) (1 pt) Fill in the missing probability for the following discrete random variable:
Ans:
X
3
6
9
10
12
P(x)
.10
.15
.25
?
.11
? = 1 - 0.61 = 0.39
(b) The number of cars per household in a small town is given by
Cars
Households
0
20
1
280
2
75
3
25
(i) (3 pts) Make a probability distribution for x where x represents the number of cars per household
in this small town.
Ans:
Cars (x)
0
1
2
3
P(x)
.0500
.7000
.1875
.0625
(ii) (5 pts) Find the mean and standard deviation for the random variable in (i)
Ans:
Mean:
 = 0(.0500)+1(.7000)+2(.1875)+3(.0625) = 1.2625
Variance:
2 = 1(.7000)+4(.1875)+9(.0625) – 1.26252 = 0.41859375
Standard Deviation:  = (0.41859375)1/2 = 0.64699
(iii) (1 pts) What is the average number of cars per household in that small town? Explain what you
mean by average.
Ans:
On average, there are 1.2625 cars per household; this is the mean number of cars per household.
This is the number one would get if they took the total number of cars in the town and divided
by the total number of households.
5. Suppose a baseball player has a batting average of 0.280 (the probability of getting a hit at an atbat). Suppose the player had 12 at-bats in a weekend series.
(a) (2 pts) What is the probability that the player had no hits?
Ans:
(.72)12  .0194
(b) (2 pts) What is the probability that the player had exactly one hit?
Ans:
C 12,1 (.28)1 (.72)11  .0906
(c) (2 pts) What is the probability that the player had 2 or more hits?
Ans:
1 - .0194 - .0906 = .8900 (This is the complementary event of 0 or 1 hits)
(d) (4 pts) Find the mean and standard deviation for the number of hits the player will get in 400 atbats.
Ans:
Mean:
 = (400)(.28) = 112
Standard Deviation:  = [(400)(.28)(.72)]1/2  8.98
6. Suppose the distribution of weights of adult male American Landrace pigs is normally distributed
with a mean of 480 lbs and standard deviation of 55 lbs.
(a) (2 pts) What weight is at the 90th percentile?
Ans:
The z-value with 90% of the normal curve to the left of it is z  1.28, thus (x-480)/55  1.28
implies x  550.4
So the 90th percentile is approximately 550.4 lbs.
(b) (2 pts) What proportion of adult male American Landrace pigs weigh between 500 lbs and 600 lbs?
Ans:
P(.36 < z < 2.18) = .9854 - .6406 = .3448
(c) (2 pts) What proportion of adult male American Landrace pigs weigh less than 600 lbs?
Ans:
P(z < 2.18) = .9854
(d) (2 pts) What proportion of adult male American Landrace pigs weigh more than 600 lbs?
Ans:
P(z > 1.82) = 1 - .9854 = .0146
(e) (2 pts) Find an interval whose center is the mean, and which contains 99% of the adult male
American Landrace pigs weights. Hint: first find a z-interval so that 99% of normal curve
lies between –z and z.
Ans:
99% of all z-values are between z = -2.575 and z = 2.575, so the interval of weights is
338.375 lbs to 621.625 lbs (the interval was found by computing 480  2.57555).
7. Alaska Airlines has found that 92% of people with tickets will show up for their Friday afternoon
flight from Seattle to Ontario. Suppose that there are 185 passengers holding tickets for this flight, and
the jet can carry 178 passengers, and that the decisions of passengers to show up are independent of
one another.
(a) (2 pts) Verify that the normal approximation of the binomial distribution can be used for this
problem.
Ans:
We check that np = 1850.92 = 170.2 > 5 and nq = 1850.08 = 14.8 > 5.
(b) (3 pts) What is the probability that 178 or fewer passengers will show up for the flight (i.e., all
passengers who show up will receive a seat)?
Ans:
We approximate the binomial distribution with the normal random variable with
 = 170.2
and
 = (1850.920.08)1/2  3.690
using the continuity correction P(r  178) = P(x < 178.5), so we compute
P(x < 178.5 ) = P(z < (178.5 – 170.2) 3.69) = P(z < 2.25) = .9878
That is, there is a 98.78% chance that all passengers will get a seat on the flight.