MAT Exam #1 September 29, 2010
Name: Answers
Directions: Read and answer each question carefully! Justify steps in proofs. Round decimals to 3 places
where applicable. Show all work! Label all probabilities!
1. As part of a quality control study aimed at improving a production line, 20 bars of soap are
weighed (in ounces). You are given the following statistics as output: Minimum = 11.6, Q1 =
14.6, Median = 16.2, Q3 = 17.4 and maximum = 20.3. The sample mean = 15.9 and sample
standard deviation = 3.71.
a. If a box-plot was constructed of the data, what outliers would there be? Justify!
b. Find the 5% trimmed mean of the data set.
IQR = 17.4- 14.6 = 2.8  1.5 IQR = 4.2
LIF = 10.4 UIF = 21.6 Since min > LIF and max < UIF there are no outliers.
x-bar = 15.9  sum = 15.9 * 20 = 318  trimmed sum = 318 – 11.6 – 20.3 = 286.1
trimmed mean = 286.1 / 18 = 15.89
2. A system contains 2 components, A and B. The system will function so long as either A or B
functions. The probability that A functions is 0.95 and the probability that B functions is 0.90.
The probability that both A and B function is 0.88.
a. Find the probability that the system functions.
b. Find the probability that only B functions.
c. Find the probability that A functions, given that B functions.
P(A) = 0.95 P(B) = 0.90
P(A ∩ B) = 0.88
P(A U B) = .95 + .90 – 0.88 = 0.97
P(A ∩B’) = .90 - .88 = .02
P(A | B) = .88/.90 = .978
3. In a bin there are 16 batteries. Of the 16, 11 work and 5 are defective. You select 4 batteries at
random.
a. Find the probability that all 4 selected batteries work.
b. Find the probability that exactly 2 selected batteries work.
c. Find the probability that at least 1 selected battery is defective.
P(4 work) = 11C4/16C4 = 330/1820 = .181
P(2 WORK) = 11C2 * 5C2 / 16C2 = 550/1820= .302
P(AT LEAST 1 D) = 1 – P(0 D) = 1 – P(4 WORK) = 1 – 330/1820 = 1490/1820 = .819
4. A new test is developed for detecting the flu. If the subject has the flu, the probability that the
test is positive is 0.95. If the subject does not have the flu, the probability the test is negative is
0.86. Assume that 20% of the population has the flu. A subject is randomly selected.
a.
Find the probability the subject has the flu, given the test is positive.
b.
Find the probability the subject does not have the flu, given the test is negative.
c.
Find the probability the subject has the flu, given the test is negative.
P(T |D) = .95
P(T’|D’) = .86
P(D) = .20
P(T’|D) = .05
P(T|D’) = .14
P(D’) = .80
P(D | T) = .95 *.20/(.95 *.20 + .14 * .80) = .19/.302 = .629
P(D’ | T’) = .86*.80/(.86 *.80 + .05 * .20) = .688/.698 = .986
P(D|T’) = 1 - .986 = .014
5. For any two events A and B, prove that if P(A|B) < P(A) then P(B|A) < P(B)
Proof.
P(A |B) < P(A)  P(A ∩ B)/P(B) < P(A) by def of Conditional Probability.
P(A ∩ B) /P(A) P(B) < P(A)/P(A) by dividing both sides by P(A) ≠0
P(A ∩ B) /P(A) P(B) < 1
P(A ∩ B) /P(A) < P(B) multiply both sides by P(B)
P(B | A) < P(B). QED
6. For any two events E and F, prove that if E and F are independent then E’ and F are
independent.
Proof.
P(E’ | F) = P(E’ ∩ F)/P(F) by def of Conditional Probability.
P(E’ | F) = [P(F) – P(E ∩ F)]/P(F) because P(F) = P(E’ ∩ F) + P(E ∩ F)
P(E’ | F) = [P(F) – (P(E)P(F)]/P(F) because E and F are Independent
P(E’ | F) = 1 – P(E)
factoring P(F) in the numerator and canceling
P(E’|F) = P(E’) by definition of compliment and therefore E’ and F are independent.
7. There are 2 bins that each contain 10 balls. Bin 1 has 7 Red and 3 Green balls. Bin 2 has 6 Red
and 4 Green balls. First the player selects a bin. The probability that the player selects Bin 1 is ¾
and the probability that the player selects Bin 2 is ¼. The player then randomly selects 2 balls
from the bin without replacement.
a. Find the probability that the two balls selected are Green given Bin 1 was selected.
b. Find the probability that the two balls selected are Green given Bin 2 was selected.
c. Find the probability that Bin 2 was selected given the 2 balls were Green.
P(2G | Bin 1) = 3C2/10C2 = 3/45 = .067
P(2G | Bin 2) = 4C2/10C2 = 6/45 = .133
P(Bin 2| 2G) = P(2G ∩ Bin 2) / P(2G) = P(2G | Bin 2) P(Bin 2) / P(2G)
P(Bin 2| 2G) = 6/45 * ¼ / (6/45 * ¼ + 3/45 * ¾) = 6/180/ 15/180 = 6/15 = 0.4