Mathematics 375 – Probability and Statistics 1 Solutions for Midterm Exam 2a November 19, 2005

I. (10) In a shipment of 14 television sets, 5 have defective “on” switches and will not power up. Out of the shipment, 7 sets are selected at random. What is the probability that at most 3 will not power up?

Solution: The number Y that will not power up is hypergeometric with N = 14, n = 7, r = 5. So

P (Y ≤ 3) = \frac{\binom{5}{0}\binom{9}{7} + \binom{5}{1}\binom{9}{6} + \binom{5}{2}\binom{9}{5} + \binom{5}{3}\binom{9}{4}}{\binom{14}{7}}

II. A) (10) The number of defects Y in any 1-foot segment of a magnetic tape is a Poisson

random variable with mean 5. What is the probability that a 1-foot segment contains exactly 4 defects?

Solution: λ = 5 in the Poisson formula. The probability is \frac{5^{4}e^{−5}}{4!} ≐ .175.

B) (15) Show, by any applicable method, that if number of defects in one 1-foot segment is Y_{1} and the number in a second 1-foot segment (not overlapping the first) is Y_{2}, then the total number of defects, Y = Y_{1} + Y_{2}, has a Poisson distribution with mean 10. (Assume defects in the two segments are independent.)

Solution: We show in general that if Y_{1} is Poisson with mean λ_{1}, Y_{2} is Poisson with mean λ_{2}, and Y_{i} are independent, then Y = Y_{1} + Y_{2} also has a Poisson distribution, with mean λ = λ_{1} + λ_{2}. One method for this is to actually work out the form of the probability mass function for Y. Note that we need to add up all the ways to get a sum of y for Y:

P (Y = y) = ∑_{y_{1}=0}^{y} P (Y_{1} = y_{1})P (Y_{2} = y − y_{1})

= ∑_{y_{1}=0}^{y} \frac{λ_{1}^{y_{1}}e^{−y_{1}}}{y_{1}!} \frac{λ_{2}^{y−y_{1}}e^{−y_{2}}}{(y − y_{1})!}

= \frac{e^{−(λ_{1}+λ_{2})}}{y!} ∑_{y_{1}=0}^{y} \binom{y}{y_{1}}λ_{1}^{y_{1}}λ_{2}^{y−y_{1}}

= \frac{e^{−(λ_{1}+λ_{2})}}{y!} (λ_{1} + λ_{2})^{y}

(where we used the binomial theorem at the last step). This last line is exactly the form of the Poisson probability mass function with λ = λ_{1} + λ_{2}. So we are done. (Alternately, we could apply the method of moment generating functions as we discussed in class(!))

III. In a biomedical research experiment, the survival time Y (in weeks) of an animal exposed to a certain fatal disease is a random variable with a gamma distribution with α = 3 and β = 2. A) (5) What is the mean survival time of a randomly selected exposed animal?

Solution: E(Y) = αβ = 6. For future reference, note that V (Y) = αβ^{2} = 12.

B) (10) Use Tchebysheff’s Theorem to estimate the probability that a single exposed animal will survive longer than 29 weeks.

Solution: We want to use the statement

P (|Y − μ| > kσ) = P (Y > μ + kσ) + P (Y < μ − kσ) ≤ \frac{1}{k^{2}}

to estimate P (Y > 29). This means that we take μ + kσ = 29, so 6 + k\sqrt{12} = 29, or k ≐ 6.639. This gives P (Y > 29) < \frac{1}{(6.639)^{2}} = .023.

C) (15) If 6 exposed animals are chosen at random, use your answer from part B to estimate the probability at least 2 of them survive longer than 29 weeks.

Solution: The number N that survive is a binomial random variable, so

P (N ≥ 2) = ∑_{n=2}^{6} \binom{6}{n}(.023)^{n}(.977)^{6−n}

D) (15) The cost of treating an exposed animal is C = 5 + .2Y^{2}. Find the mean and variance of C.

Solution: First

E(C) = 5 + (.2)E(Y^{2}) = 5 + (.2)(V (Y) + (E(Y))^{2}) = 5 + (.2)(12 + 36) = 14.6.

Then to compute the variance of the cost:

E(C^{2}) = ∫_{0}^{∞}(5 + .2y^{2})^{2} \frac{y^{2}e^{−y/2}}{Γ(3) · 8} dy

= \frac{1}{16} (25 ∫_{0}^{∞} y^{2}e^{−y/2} dy + 2 ∫_{0}^{∞} y^{4}e^{−y/2} dy + (.04) ∫_{0}^{∞} y^{6}e^{−y/2} dy)

= \frac{1}{16} (25 · 2^{3}Γ(3) + 2 · 2^{5}Γ(5) + (.04) · 2^{7}Γ(7))

= 351.4

Then V (C) = E(C^{2}) − (E(C))^{2} = 351.4 − (14.6)^{2} ≐ 138.24.

IV. A) (10) The lifetime of a certain type of small motor is uniformly distributed on the

interval [0, 30]. What is the probability that the motor will last more than 10 years?

Solution: P (Y > 10) = ∫_{10}^{30} \frac{dy}{30−0} = \frac{2}{3}.

B) (10) Now, assume the lifetime of the motors is normally distributed with mean 15 years and standard deviation 4 years. The manufacturer replaces free all motors that fail while under warrantee. But to avoid going out of business because of the cost of doing the free replacements, the manufacturer must set the warrantee so that they will have to replace no more than 3% of the motors that fail. What should the length of the warrantee be?

Solution: Let w be the length of the warrantee. Then we want P (Y < w) = .03. After standardizing, this says P (Z < \frac{w−15}{4}) = .03. From the standard normal table, this says \frac{w−15}{4} = −1.88, so w = 15 − (4)(1.88) = 7.48 years.

Extra Credit (10) Let Z be a standard normal random variable. What is the momentgenerating function for Y = 8Z − 4.3? What is the distribution of Y?

Solution:

m_{Y} (t) = e^{−4.3t}m_{Z}(8t) = e^{−4.3t+ \frac{64t^{2}}{2}} .

By the Uniqueness Theorem, Y has a normal distribution too, with μ = −4.3 and σ = 8.