Mathematics 375 – Probability and Statistics I Solutions for Midterm Exam 1 – October 8, 2009

I. To test the effectiveness of a seal for air leaks in automobile tires, after the seal was installed in a tire, a needle was inserted into the tire and air pressure was increased until leakage was observed. The pressures (in lb per square inch) where leakage first occurred on 10 trials were:

A) (10) Construct a relative frequency histogram for this data on the interval [86.5, 96.5], subdividing into 5 equal “bins”

(86.5, 88.5], (88.5, 90.5], (90.5, 92.5], (92.5, 94.5], (94.5, 96.5].

Solution: The histogram should have a box of height 2/10 on the first interval, a box of height 2/10 on the second interval, a box of height 3/10 on the third interval, a box of height 1/10 on the fourth interval, and a box of height 2/10 on the last interval.

C) (15) How many of the data points are within 1 standard deviation of the sample mean?

Solution: The sample mean is

\overline{x} = \frac{93 + 87 + 91 + 90 + 92 + 88 + 95 + 91 + 90 + 96}{10} = 91.3.

The sample variance is

s^{2} = \frac{1}{9} ∑_{i=1}^{10}(x_{i} − \overline{x})^{2} ≐ 8.01.

So the sample standard deviation is s = \sqrt{8.01} ≐ 2.83. The interval [\overline{x} − s, \overline{x} + s] = [88.47, 94.13]. Six of the 10 data values are in this interval. (Note: This is relatively close to the empirical rule of 68%.)

II. Sailors on a ship can send signal by arranging (exactly) 8 colored flags along a rope line. A) (5) How many different signals can they send if they have flags of 8 different colors?

Solution: This is the number of permutations of 8 things, 8 at a time: 8! (factorial, not exclamation point).

B) (10) How many different signals can they send if they have 4 red, 2 green, and 2 blue flags? (There are no differences between the flags of the same color.)

Solution: The red flags can occur in any 4 of the 8 positions along the rope line. Once the reds are placed, the greens go in 2 of the remaining 4, and then the blues go in the remaining two slots. By the “m · n rule,” this gives

\binom{8}{4} · \binom{4}{2} · \binom{2}{2} = 420

possible arrangements. Note: this is the same as the multinomial coefficient

\binom{8}{4 2 2} .

C) (10) In the situation of part B, if a random arrangement of flags is constructed, what is the probability that at least 5 other flags appear between the two blue flags?

Solution: The answer from part B will be the denominator of the ratio giving the probability we want to compute. The numerator is the number of arrangements of the 8 flags where the two blues are separated by at least 5 other flags. Note that the possible locations for the blue flags are as shown:

BXXXXXBX, BXXXXXXB, or XBXXXXXB.

In each case, the red and green flags can be placed in \binom{6}{4}\binom{2}{2} ways (reasoning as in part B). Therefore the probability we want is

\frac{3\binom{6}{4}\binom{2}{2}}{\binom{8}{4} · \binom{4}{2} · \binom{2}{2}} = \frac{3\binom{6}{4}}{\binom{8}{4}\binom{4}{2}}.

III. Let A_{1}, A_{2}, A_{3} be events in a sample space S. Assume that S = A_{1} ∪ A_{2} ∪ A_{3}, where A_{i} ∩ A_{j} = ∅ if i ≠ j. Let P (A_{1}) = .35, P (A_{2}) = .4, P (A_{3}) = .25. Finally, let B be another event with P (B|A_{1}) = .1, P (B|A_{2}) = .2 and P (B|A_{3}) = .05. A) (10) State the Law of Total Probability and use it to compute P (B).

The LTP says that if we have a partition S = A_{1}∪⋯∪A_{k} into events with A_{i}∩A_{j} = ∅ whenever i ≠ j, then for all events B,

P (B) = P (B|A_{1})P (A_{1}) + ⋯ + P (B|A_{k})P (A_{k}).

Applying this here:

P (B) = (.1)(.35) + (.2)(.4) + (.05)(.25) = .1275.

B) (5) Are B and A_{2} independent events? Why or why not?

Solution: No. The events B and A_{2} are not independent, since P (B ∩ A_{2}) = P (B|A_{2})P (A_{2}) = .08 is not the same as P (B)P (A_{2}) = (.1275)(.4) = .051.

C) (10) What is P (A_{1} ∪ A_{2}|B)? State the name(s) of any general rule(s) you are using.

Solution: Using the definition of conditional probabilities, P (A_{1}∪A_{2}|B) = \frac{P ((A_{1}∪A_{2})∩B)}{P (B)}. Using standard properties of set operations, (A_{1} ∪ A_{2}) ∩ B = (A_{1} ∩ B) ∪ (A_{2} ∩ B). Since A_{1} ∩ A_{2} = ∅, it follows that (A_{1} ∩ B) ∩ (A_{2} ∩ B) = ∅ also. Therefore, P ((A_{1} ∪ A_{2}) ∩ B) = P (A_{1} ∩ B) + P (A_{2} ∩ B) by the additive rule. It follows by Bayes’ Rule that the probability we want is

\frac{P (B|A_{1})P (A_{1}) + P (B|A_{2})P (A_{2})}{P (B)} = 1 − \frac{P (B|A_{3})}{P (B)} = 1 − (.098) ≐ .902.

IV. A) (10) Let Y have a binomial distribution with parameters n = number of trials and

p = success probability on each trial. Show that E(Y) = np.

Solution: See the class notes or the text.

B) (15) An army regiment has 20 different squads, 16 with 25 soldiers each, 3 with 100 soldiers each, and one with 300 soldiers, for a total of 1000 soldiers. Select a soldier at random out of the 1000 and let the random variable X equal the size of the squad to which that soldier belongs. Find the probability mass function, the expected value, and the variance of X.

Solution: The possible values with nonzero probabilities are x = 25, 100, 300. The probability that X takes one of these values is the probability that the randomly selected soldier belongs to a squad of that size:

P (X = 25) = \frac{400}{1000} = .4
P (X = 100) = \frac{300}{1000} = .3
P (X = 300) = \frac{300}{1000} = .3
P (X = x) = 0 for all other x.

The expected value is E(X) = (25)(.4) + (100)(.3) + (300)(.3) = 130. The variance is

V (X) = E((X −130)^{2}) = (25−130)^{2}(.4)+(100−130)^{2}(.3)+(300−130)^{2}(.3) = 13350.

(The standard deviation would be more meaningful here: σ = \sqrt{13350} ≐ 115.5.

Extra Credit (10)

Let X be a discrete random variable which takes as values all natural numbers k ≥ 1. Assuming that E(X) exists, show that

E(X) = ∑_{k=1}^{∞} P (X ≥ k).

(Also, why do we need to add the hypothesis that E(X) exists in a case like this?)

Solution: We start from the definition of the expected value and then regroup the terms in the sum (this is justified since the infinite series must be absolutely convergent – it has all non-negative terms!) :

E(X) = ∑_{k=1}^{∞} kP (X = k)

= P (X = 1) + 2P (x = 2) + 3P (X = 3) + ⋯ = P (X = 1) + P (X = 2) + P (X = 3) + ⋯

+ P (X = 2) + P (X = 3) + ⋯
+ P (X = 3) + ⋯
+ ⋯

= P (X ≥ 1) + P (X ≥ 2) + P (X ≥ 3) + ⋯

= ∑_{k=1}^{∞} P (X ≥ k).

The hypothesis on E(X) is necessary for this kind of random variable since for some other Y, ∑_{y=1}^{∞} P (Y = y) would exist and sum to 1, but the series ∑_{y=1}^{∞} yP (Y = y) might diverge. An example would be a random variable with

P (Y = y) = \frac{6}{π^{2}} \frac{1}{y^{2}} .

The series ∑_{y=1}^{∞} \frac{1}{y^{2}} converges (p-series with p = 2). The sum is actually \frac{π^{2}}{6} , and that accounts for the normalization factor in the formula for P (Y = y) above. But ∑_{y=1} y \frac{1}{y^{2}} = ∑_{y=1}^{∞} \frac{1}{y} diverges (harmonic series).