[Latexpage]
Answers and hints or partial solutions are below the fold… up to #5 for now, to be updated this evening…
1) This is negative binomial (waiting time for binomial success). To compute it from the binomial probability distribution, use the useful formula $b*(x; k, p) = \frac{k}{x}b(k; x, p)$
$b*(5; 1, 0.2) = \frac{1}{5}b(1; 5, 0.2) = 0.08192$
2a) This is Poisson with $lambda = 3$: $P(X < 2) = p(0, 3) + p(1, 3) \approx 0.1991$
2b) This is waiting time for Poisson, namely, it has exponential density $f(x) = 3e^{-3x}$
Also, the unit of time is 15 minutes, so 6 minutes is $\frac{6}{15} = \frac{2}{5}$ of a unit of time.
$P(X > 6) = \displaystyle \int_{2/5}^{\infty}3e^{-3x}\mathrm{d}x = e^{-1.2} \approx 0.6988$
3a) This is binomial with n = 20 and p = 0.05
$P(X \le 1) = b(0; 20, 0.05) + b(1; 20, 0.05) \approx 0.7358$
3b) n must be “large” and p must be “small”. According to the source linked in the Wikipedia article on Poisson distribution :
“There is a rule of thumb stating that the Poisson distribution is a good approximation of the binomial distribution if n is at least 20 and p is smaller than or equal to 0.05, and an excellent approximation if n ≥ 100 and np ≤ 10.”
So our example passes (barely) by the first rule of thumb but not the second. We can use the Poisson distribution.
3c) $lambda = np = 1$
P(X \le 1) = p(0; 1) + p(1; 1) \approx 0.7358$
3d) Percentage error = (estimated value – true value)/true value , translated into percents.
Here the percentage error (using more decimal places of the computations above) is $\frac{0.7357589 – 0.7358395}{0.7358395} \approx -0.0001$ or 0.01%
3e) We need both np and n(1-p) to be at least 5, or preferably at least 10.
Here np =1, so we cannot use the normal approximation. But we go ahead and do it anyway and see what happens!
3f) Estimate by the normal probability with $\mu = np = 1$ and $\sigma = \sqrt{np(1-p)} = \sqrt{0.95}$. Don’t forget the continuity correction
$P(x \le 1) \approx n(X \le 1.5; 1, \sqrt{0.95}) \approx 0.7006656$, not very close!
3g) $\frac{0.7006656 – 0.7358395}{0.7358395} \approx -0.048$ or 4.8%, a fairly big error! (That is because we did not meet the conditions for using this approximation.)
4a) $\Gamma(7) = (7-1)! = 6! = 720$
4b) Use the recurrence relation $\Gamma(n+1) = n\Gamma(n)$, since $\frac{3}{2} = \frac{1}{2} + 1$
$\Gamma\left(\frac{3}{2}\right) = \frac{1}{2}\Gamma\left(\frac{1}{2}\right) = \frac{\sqrt{\p}i}{2}$
This was a homework problem!
[In fact it is true that $\Gamma\left(\frac{1}{2}\right) = \sqrt{\pi}}$, which mean that you could say that $\left(-\frac{1}{2}\right)! = \sqrt{\pi}}$ by the rule that $\Gamma{n} = (n-1)!$]
5) $Z_{0.025} \approx 1.96$
Make sure that you know how to compute these critical values: don’t just memorize this one!
6a) n(X< 7.50; 8, 0.25) $\approx$ 0.0228
6b) n(7.5 < X < 8.5; 8, 0.25) $\approx$ 0.9549
6c) $n(\bar{X} < 7.9; 8, \frac{0.25}{\sqrt{25}) \approx 0.0228
7) $Z_{0.01} \approx 2.33$
The margin of error is $\approx 2.33\cdot \frac{0.3}{\sqrt{50}} \approx 0.09$
confidence interval: (3.91, 4.09)
