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

I. (10) In a group of 12 chickens, 5 are infected with the H5N1 “bird flu” virus and the others are not. If 7 chickens are selected randomly from the group, what is the probability 2 or more have the virus?

Solution: The number Y that do have the virus is hypergeometric with N = 12, n = 7, r = 5. So

P (Y ≥ 2) = \frac{\binom{5}{2}\binom{7}{5} + \binom{5}{3}\binom{7}{4} + \binom{5}{4}\binom{7}{3} + \binom{5}{5}\binom{7}{3}}{\binom{12}{7}}

II. A) (15) Assume that Y_{1} has a Poisson distribution with mean λ_{1} and Y_{2} has Poisson

distribution with mean λ_{2}. Also assume that Y_{1} = y_{1} and Y_{2} = y_{2} are independent events for all y_{1} ≥ 0 and y_{2} ≥ 0. Show, by any applicable method, that Y = Y_{1} + Y_{2} also has a Poisson distribution, with mean λ = λ_{1} + λ_{2}.

Solution: 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(!))

B) (10) The number of defects Y in any 2-foot segment of a magnetic tape is a Poisson random variable with mean 3. Use part A to find the probability that a 4-foot segment contains exactly 8 defects, assuming the necessary independence. (You can use part A even if you didn’t completely see how to do that part.)

Solution: The number of defects in the 4-foot segment is Poisson with λ = 6 by part A. Hence the probability is P (Y = 8) = \frac{6^{8}e^{−6}}{8!} ≐ .103.

A) (10) The lifetime of a certain type of battery is uniformly distributed on the interval [0, 25] (years). What is the probability that the battery will last more than 10 years?

Solution: P (Y > 10) = ∫_{10}^{25} \frac{dy}{25−0} = \frac{3}{5}.

B) (10) Now, assume the lifetime of the batteries is normally distributed with mean 10 years and standard deviation 4 years. The manufacturer replaces free all batteries 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 5% of the batteries 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) = .05. After standardizing, this says P (Z < \frac{w−10}{4}) = .05. From the standard normal table, this says \frac{w−10}{4} = −1.645, so w = 10 − (4)(1.645) = 3.42 years.

IV. 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 = 4 and β = 3. A) (5) What is the mean survival time of a randomly selected exposed animal?

Solution: E(Y) = αβ = 12. For future reference, note that V (Y) = αβ^{2} = 36.

B) (10) Use Tchebysheff’s Theorem to estimate the probability that a single exposed animal will survive longer than 30 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 > 30). This means that we take μ + kσ = 30, so 12 + 6k = 30, or k = 3. This gives P (Y > 30) < \frac{1}{3^{2}} = .111.

C) (15) If 6 exposed animals are chosen at random, use your answer from part B to estimate the probability no more than 2 of them survive longer than 30 weeks.

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

P (N ≤ 2) = ∑_{n=0}^{2} \binom{6}{n}(.111)^{n}(.889)^{6−n}

D) (15) The cost of treating an exposed animal is C = 4 + .25Y^{2}. Find the mean and variance of C.

Solution: First E(C) = 4 + (.25)E(Y^{2}) = 4 + (.25)(V (Y) + (E(Y))^{2}) = 4 + (.25)(36 + 144) = 49. Then to compute the variance of the cost:

E(C^{2}) = E(16 + 2Y^{2} + \frac{1}{16}Y^{4})

= 16 + 2E(Y^{2}) + \frac{1}{16}E(Y^{4})

= 16 + 2(180) + \frac{1}{16 · Γ(4) · 81} ∫_{0}^{∞} y^{7}e^{−y/3} dy

= 376 + \frac{1}{96 · 81} · 3^{8}Γ(8)

= 4628.5

Then V (C) = E(C^{2}) − (E(C))^{2} = 4628.5 − 49^{2} ≐ 2227.5.

Extra Credit (10) Let Z be a standard normal random variable. What is the momentgenerating function for Y = 3Z + 5? What is the distribution of Y?

Solution:

m_{Y} (t) = e^{5t}m_{Z}(3t) = e^{5t+ \frac{9t^{2}}{2}} .

By the Uniqueness Theorem, Y has a normal distribution too, with μ = 5 and σ = 3.