Mathematics 375 – Probability Theory Solutions – Midterm Exam 2 Practice Problems November 9, 2011

I. If Y denotes the number of underage students who get carded, then Y is hypergeometric with N = 10, r = 4, n = 5. So

P (Y = 2) = \begin{matrix}\binom{4}{2} & \binom{6}{3} \\ ( & {10 \atop 5} & ) \end{matrix} = \frac{10}{21} ≐ .48

II. By examining the form of the density function, we see that Y has a beta distribution with α = β = 3. Hence: A) The constant c must be

c = \frac{1}{B(3, 3)} = \frac{Γ(6)}{Γ(3)Γ(3)} = \frac{5!}{2!2!} = 30

B) The variance is

V (Y) = \frac{αβ}{(α + β)^{2}(α + β + 1)} = \frac{3 · 3}{36 · 7} = \frac{1}{28}

C) The cumulative distribution function is the antiderivative F (y) of the density with F (−∞) = 0 and F (+∞) = 1. This is

F (y) = \{ \begin{matrix}0 & if y < 0 \\ 10y^{3} − 15y^{4} + 6y^{5} & if 0 ≤ y < 1 \\ 1 & if y ≥ 1 \end{matrix}

D) This is

∫_{.1}^{.25} 30y^{2}(1 − y)^{2} dy = F (.25) − F (.1) ≐ .09496.

III. The lifetime Y of a single switch has the exponential density

f(y) = \{ \begin{matrix}\frac{1}{2}e^{−y/2} & if y > 0 \\ 0 & otherwise \end{matrix}

A) The probability that a single switch fails during the first year is the same as the probability that its life is less than 1:

P (Y < 1) = ∫_{0}^{1} \frac{1}{2}e^{−y/2} dy = −e^{−y/2}|_{0}^{1} = 1 − e^{−1/2} ≐ .3935

B) The probability that at most 30 out of 100 of these switches will fail in the first year is computed by the binomial probability formula (since we assume the switches operate independently):

∑_{y=0}^{30} \binom{100}{y}(.3935)^{y}(.6065)^{100−y}.

IV. A) Let Y be the weight of one poodle

P (7.3 < Y < 9.1) = P (\frac{7.3 − 8}{.9} < \frac{Y − 8}{.9} < \frac{9.1 − 8}{.9})

This is the same as

P (−.78 < \frac{Y − 8}{.9} < 1.22)

We know Z = \frac{Y −8}{.9} is a standard normal, so we can use the standard normal table to find this probability. By the symmetry of the normal density,

P (−.78 < Z < 0) = P (0 < Z < .78) = .5 − .2177 = .2823

We also have

P (0 < Z < 1.22) = .5 − .1112 = .3888

Hence

P (−.78 < Z < 1.22) = .2823 + .3888 = .6711

B) Partial credit would be given for an answer like

E(Y^{3}) = ∫_{−∞}^{∞} y^{3} \frac{1}{\sqrt{2π(.9)^{2}}} e^{−(y−8)^{2}/2(.9)^{2}} dy.

But the only reasonable way to get a numerical value for this is to think of using the moment-generating function for Y. E(Y^{3}) = m′′′(0) where m(t) is the mgf for a normal random variable. The general form of this is

m(t) = e^{\frac{σ^{2}t}{2} +μt}

Hence differentiating with the chain and product rules:

\begin{aligned}m′(t) & = e^{\frac{σ^{2}t}{2} +μt}(σ^{2}t + μ) \\ m′′(t) & = e^{\frac{σ^{2}t}{2} +μt}((σ^{2}t + μ)^{2} + σ^{2}) \\ m′′′(t) & = e^{\frac{σ^{2}t}{2} +μt}((σ^{2}t + μ)^{3} + 3σ^{2}(σ^{2}t + μ)) \\ ⇒ m′′′(0) & = μ^{3} + 3σ^{2}μ \end{aligned}

With μ = 8 and σ = .9, this yields E(Y^{3}) = 531.44.

V. This is the density for a Gamma-distributed random variable with α = 2 and β = 1. Hence

m(t) = (1 − t)^{−2}

(This can also be computed directly as

E(e^{tY}) = ∫_{0}^{∞} e^{ty} · ye^{−y} dy.)

VI. From the situation, it is clear that the selection of missiles is being done without replacement. Hence the required probabilities will be computed using the hypergeometric formulas with N = 10, r = 3, n = 4. Let Y be the number that will not fire out of the four selected A) The probability that all four of the chosen missiles will fire is

P (Y = 0) = \begin{matrix}\binom{3}{0} & \binom{7}{4} \\ ( & {10 \atop 4} & ) \end{matrix} = \frac{1}{6}

B) The probability that at most 2 will not fire is

P (Y ≤ 2) = \frac{\binom{3}{2}\binom{7}{2} + \binom{3}{1}\binom{7}{3} + \binom{3}{0}\binom{7}{4}}{\binom{10}{4}} = \frac{29}{30}.

VII. A) The mean survival time is E(Y) = αβ = 12 weeks. B) This is not possible since 12\sqrt{2} = 2σ. By Tchebysheff’s Theorem, P (|Y − μ| < 2σ) ≥

.75 for every random variable (independent of its distribution). C) The probability that a single animal survives 15 or more weeks is computed like this

(integrating by parts):

p = ∫_{15}^{∞} \frac{ye^{−y/6}}{36} dy

= \frac{−1}{6}ye^{−y/6}|_{15}^{∞} + \frac{1}{6} ∫_{15}^{∞} e^{−y/6} dy

= \frac{5}{2}e^{−5/2} + e^{−5/2}

= \frac{7}{2}e^{−5/2} ≐ .2873

Then if Y is the number out of the 6 that survive longer than 15 weeks,

P (Y ≥ 2) = 1 − P (Y = 0) − P (Y = 1) = 1 − \binom{6}{0}p^{0}(1 − p)^{6} − \binom{6}{1}p^{1}(1 − p)^{5} ≐ .5520.

(Note: The original posting of this solution contained an error because the Gamma density used did not incorporate the proper value of β = 6.)

VIII. Let w be the warrantee period and Y be the lifetime of a randomly chosen motor. We want to choose w so that the probability that a motor lasts less than the warrantee period is only .03, or in other words: P (Y < w) = .03. Since Y ∼ Normal(10, 4), this is the same as

.03 = P (Y < w) = P (Z < \frac{w − 10}{2})

From the standard normal table, we see \frac{w−10}{2} ≐ −1.88 so w = 6.24 years.

IX. The joint density is nonzero on the triangle in the y_{1}, y_{2} plane with vertices at (0, −1), (1, 0), and (0, 1). A) Then

P (Y_{2} > 0) = ∫_{0}^{1} ∫_{0}^{1−y_{1}} 30y_{1}y_{2}^{2} dy_{2} dy_{1} = \frac{1}{2}

(This can be seen without calculation if you notice the symmetry of the region and the density function under reflection across the y_{1}-axis; otherwise just compute!)

B) The marginal density of Y_{1} is 0 for y_{1} ∉ [0, 1] and

∫_{y_{1}−1}^{1−y_{1}} 30y_{1}y_{2}^{2} dy_{2} = 20y_{1}(1 − y_{1})^{3}

if 0 ≤ y_{1} ≤ 1. Since

B(2, 5) = \frac{Γ(2)Γ(4)}{Γ(6)} = \frac{3!}{5} = \frac{1}{20}

This is the beta density with α = 2 and β = 4. C) The conditional density is computed as f(y_{1}, y_{2})/f_{1}(y_{1}). So by part B), this gives

\begin{aligned}f(Y_{2}|Y_{1} = y_{1}) & = \frac{30y_{1}y_{2}^{2}}{20y_{1}(1 − y_{1})^{3}} \\ & = \frac{3y_{2}^{2}}{2(1 − y_{1})^{3}} \end{aligned}

for −1 ≤ y_{2} ≤ 1 and zero otherwise. (Note: we would use this with a particular value of y_{1} substituted in, and think of it as a function of y_{2}.)