Mathematics 376 – Probability and Statistics II Problem Set 2 Solutions February 6, 2004

7.23/Since σ = 2.5 and n = 100, by the Central Limit Theorem we have

P (−.5 ≤ \overline{Y} − μ ≤ .5) = P (10 · \frac{−.5}{2.5} ≤ 10 · \frac{\overline{Y} − μ}{2.5} ≤ 10 · \frac{.5}{2.5}) ≐ P (−2 ≤ Z ≤ 2)

(where Z is standard normal – note that n = 100 is in the range where we expect

\sqrt{n} · \frac{\overline{Y} − μ}{σ}

to be approximately normal). Hence from the standard normal table, the desired probability is .9544.

7.24/Continuing from the setup of the previous problem, to get

P (\frac{(−.4)\sqrt{n}}{2.5} ≤ \sqrt{n} · \frac{\overline{Y} − μ}{2.5} ≤ \frac{(.4)\sqrt{n}}{2.5}) = .95

we treat \sqrt{n} · \frac{\overline{Y} −μ}{2.5} as approximately standard normal. Then from the normal table we want \frac{(.4)\sqrt{n}}{2.5} = 1.96, so n = 151 (rounding up to the next largest whole number.

7.25/Since the size of the ethnic group, n = 64, is relatively large, we can answer the given question by finding the probability that a randomly selected sample of size n = 64 has mean wage less than or equal to $6.90, applying the Central Limit Theorem:

P (\overline{Y} ≤ 6.90) ≐ P (Z ≤ \sqrt{64} · \frac{6.90 − 7.00}{.5}) = P (Z ≤ −1.6) = .0548

In other words, there is about a 5.5% chance that a random sample of size n = 64 will have mean wage that small. In other words, it’s possible, but quite unlikely, that the workers from this ethnic group are being treated fairly.

7.33/ a) No, it does not seem that the concentrations are normally distributed. With a normal distribution and μ = 12 and σ = 9, there would be a significant chance that the value would be negative (P (Y < 0) = P (Z < −1.33) = .0918). But on physical grounds, negative values are not possible for concentrations. b) As in the previous problems, n = 100 is large enough that we expect the Central Limit Theorem to have “kicked in”:

P (\overline{Y} > 14) ≐ P (Z ≥ 10 · \frac{14 − 12}{9}) = .0132.

7.37/Each lamp has mean life μ = 50 and standard deviation σ = 4. There are 25 lamps in all, so writing Y_{i} for the life of the ith lamp, we are looking for

P (∑_{i=1}^{2} Y_{i} > 1300) = P (\overline{Y} > 1300/25) = P (\overline{Y} > 52)

As in the previous problems, we will assume that n is large enough so that the Central Limit Theorem applies (with somewhat less “ironclad” justification this time!):

P (\overline{Y} > 52) = P (Z > \sqrt{25} · \frac{52 − 50}{4}) = P (Z > 2.5) = .0062,

by the standard normal table.

7.38/Let W_{i} = X_{i} − Y_{i}. Then the W_{i} are independent and identically distributed with

E(W_{i}) = μ_{1} − μ_{2} V (W_{i}) = σ_{1}^{2} + σ_{2}^{2}

Thus the Central Limit Theorem applies and we get

U_{n} = \frac{\frac{1}{n} ∑_{i=1}^{n}(X_{i} − Y_{i}) − (μ_{1} − μ_{2})}{\sqrt{\frac{σ_{1}^{2}+σ_{2}^{2}}{n}}} = \sqrt{n} (\frac{\overline{X} − \overline{Y} − (μ_{1} − μ_{2})}{\sqrt{σ_{1}^{2} + σ_{2}^{2}}})

converges to N(0, 1) in distribution as n → ∞.

7.74/From last semester, the moment generating function of a Poisson random variable Y with parameter λ has the form

m_{Y} (t) = exp(λ(e^{t} − 1))

Hence the moment generating function of Z = (Y − λ)/\sqrt{λ} = Y/\sqrt{λ} − \sqrt{λ} is (using the formula m_{aY +b}(t) = e^{bt}m_{Y} (at)):

m_{Z}(t) = exp(−t\sqrt{λ})m_{Y} (t/\sqrt{λ})
= exp(−t\sqrt{λ}) exp(λ(e^{t/\sqrt{λ}} − 1))
= exp(λe^{t/\sqrt{λ}} − λ − t\sqrt{λ})

Now we use the Taylor series for the exponential to expand the “inside” term

e^{t\sqrt{λ}} = 1 + t/\sqrt{λ} + t^{2}/(2λ) + t^{3}/(6λ^{3/2}) + ⋯

Substituting this into the above, the first two terms cancel out, and we are left with:

m_{Z}(t) = exp(t^{2}/2 + t^{3}/(6λ^{1/2}) + ⋯)

Hence (since all terms after the first have powers of λ in the denominator):

lim_{λ→∞} m_{Z}(t) = exp(t^{2}/2)

which is the mgf for a standard normal. This shows that Z → N(0, 1) in distribution as λ → ∞.