Mathematics 375 – Probability and Statistics 1 Solutions for Review Problems, Exam 2 November 18, 2003

3.157 a) Let Y be the number of imperfections. Using the formula for Poisson probabilities, we get with λ = 4

P (Y ≥ 1) = ∑_{y=1}^{∞} \frac{4^{y}}{y!} e^{−4} = 1 − e^{−4} ≐ .982

For b), the idea is the same but we take λ = 12

P (Y ≥ 1) = ∑_{y=1}^{∞} \frac{12^{y}}{y!} e^{−12} = 1 − e^{−12} ≐ .999994

3.158/Here in an 8 square yard bolt, the expected number of imperfections is λ = 32. So E(C) = E(10Y) = 10E(Y) = 10·32 = 320. V (C) = V (10Y) = 10^{2}V (Y) = 100·32 = 3200 so σ = \sqrt{3200} = 56.57. (We are using the formulas E(Y) = V (Y) = λ for a Poisson random variable here.)

3.160/We have

m(t) = (pe^{t} + q)^{n} = (p (1 + t + \frac{t^{2}}{2} + ⋯) + q)^{n} = (1 + pt + \frac{pt^{2}}{2} + ⋯)^{n}

Expanding and keeping only terms with t and t^{2} (all we need for the problem), we get:

= 1 + npt + (\binom{n}{2}p^{2} + \frac{n}{2}p) t^{2} + ⋯

Hence μ′_{1} = m′(0) = np and μ′_{2} = m′′(0) = n(n − 1)p^{2} + np. (This agrees with calculations we did in class for the binomial random variables!)

3.173 a) Hypergeometric with N = 100, r = 40 (the defectives), and n = 20, so

p(10) = \begin{matrix}\binom{40}{10} & \binom{60}{10} \\ ( & {100 \atop 20} & ) \end{matrix} ≐ .119

(Note: The approximate value would be obtained by expressing the binomial coefficients as factorials, cancelling as many common factors as you can, then carefully multiplying everything out – straightforward but very tedious. On an exam, of course, the binomial coefficient form would be acceptable!!) b) Using the binomial distribution (treating the selection of the sample of 20 as independent events each with a probability p = 40/100 = .4 of getting a defective),

p(10) ≐ \binom{20}{10}(.4)^{10}(.6)^{1}0 ≐ .117

Note that here the binomial probability is a good approximation to the hypergeometric probability (the exact value). The idea is that N = 40 is large enough so that the probabilities of getting a defective do not change that much in successive trials.

3.176/ This was done in class – see the notes.

3.178/We also did calculations like this a number of times: Cancel factors from the y(y−1) against the denominator in \binom{r}{y}, reindex the sum, use 176 c, then cancel:

E(Y (Y − 1)) = ∑_{y=0}^{n} y(y − 1)\begin{matrix}\binom{r}{y} & \binom{N−r}{n−y} \\ ( & {N \atop n} & ) \end{matrix}

= ∑_{y=2}^{n} \begin{matrix}\binom{r}{y−2} & \binom{N−r}{n−y} \\ ( & {N \atop n} & ) \end{matrix}

= ∑_{z=0}^{n−2} \frac{r(r − 1)\binom{r−2}{z}\binom{N−2−(r−2)}{(n−2)−z}}{\binom{N}{n}}

= \frac{r(r − 1)\binom{N−2}{n−2}}{\binom{N}{n}}

= \frac{r(r − 1)n(n − 1)}{N(N − 1)}

4.67/See class notes.

4.88/By inspection, this has a gamma distribution with α = 3, β = 1/2. So E(Y) = αβ = 3/2 and V (Y) = αβ^{2} = 3/4.

4.101/See solutions for Problem Set 7.

4.110/a) Using the general formula for a uniformly distributed random variable, with θ_{1} = 0 and θ_{2} = 1, we have

m_{Y} (t) = \frac{e^{t} − 1}{t}

For the next part we use the formula m_{(}aY + b)(t) = e^{b}tm_{Y} (at) proved on Problem Set 8.

b) We get:

m_{aY} (t) = \frac{e^{at} − 1}{at}

This is the mgf for a uniform r.v. on [0, a]. c) We get:

m_{−aY} (t) = \frac{e^{−at} − 1}{−at} = \frac{1 − e^{−at}}{at}

This is the mgf for a uniform r.v. on [−a, 0]. d) We get

m_{aY +b}(t) = e^{bt} \frac{e^{at} − 1}{at} = \frac{e^{(a+b)t} − e^{bt}}{at}

This is the mgf for a uniform r.v. on [b, a + b].