Solutions

1.
To solve this problem, you need to know that the distribution of the mean of a sample of size n from a normal distribution of mean μ and standard deviation σ is normal with mean μ and standard deviation ${\frac{{\sigma }}{{\sqrt{n}}}}$. Thus, the probability is BITMAPSETProbSolvHint0.1807in0.2188in0inq1

Pr$\displaystyle \left(\vphantom{ \bar{X}\leq 46}\right.$$\displaystyle \bar{{X}}$≤46$\displaystyle \left.\vphantom{ \bar{X}\leq 46}\right)$ = $\displaystyle \func$NormalDist$\displaystyle \left(\vphantom{ 46;50,\frac{1}{2}%
\sqrt{5}}\right.$46;50,$\displaystyle {\frac{{1}}{{2}%
}}$$\displaystyle \sqrt{{5}}$$\displaystyle \left.\vphantom{ 46;50,\frac{1}{2}%
\sqrt{5}}\right)$ =  1.7331×10-4

2.
$\func$ExponentialDist$\left(\vphantom{ 7.5;5}\right.$7.5;5$\left.\vphantom{ 7.5;5}\right)$ =  .78 = P(X≤7.5), so the probability that X is greater than 7.5 is - .78 =  .22.

$\func$ExponentialDist$\left(\vphantom{ 2;5}\right.$2;5$\left.\vphantom{ 2;5}\right)$ =  .33 = P(X≤2), so the answer to the second question is ``about 33 percent.''BITMAPSETProbSolvHint0.1807in0.2188in0inq1

3.
$\func$NormalDist$\left(\vphantom{ 7.5;5,2}\right.$7.5;5, 2$\left.\vphantom{ 7.5;5,2}\right)$ =  .894 = P(X≤7.5), so the probability that X is greater than 7.5 is -  .894  =  .106, or 10.6 percent.BITMAPSETProbSolvHint0.1807in0.2188in0inq3

$\func$NormalDist$\left(\vphantom{ 2;5,2}\right.$2;5, 2$\left.\vphantom{ 2;5,2}\right)$ = .067 = P(X≤2), so the answer to the second question is ``about 7 percent.''

4.
For the normal distribution, Evaluate gives
$\displaystyle \int_{{-\infty }}^{{\infty }}$$\displaystyle \limfunc$NormalDen$\displaystyle \left(\vphantom{ u;\mu ,\sigma }\right.$u;μ, σ$\displaystyle \left.\vphantom{ u;\mu ,\sigma }\right)$udu =  μ  
$\displaystyle \int_{{-\infty }}^{{\infty }}$$\displaystyle \limfunc$NormalDen$\displaystyle \left(\vphantom{ u;\mu ,\sigma }\right.$u;μ, σ$\displaystyle \left.\vphantom{ u;\mu ,\sigma }\right)$$\displaystyle \left(\vphantom{ u-\mu }\right.$u - μ$\displaystyle \left.\vphantom{ u-\mu }\right)^{{2}}_{}$du =  σ2  

For the Student's t distribution, with five degrees of freedom, Evaluate gives

$\displaystyle \int_{{-\infty }}^{{\infty }}$u$\displaystyle \limfunc$TDen$\displaystyle \left(\vphantom{ u;5}\right.$u;5$\displaystyle \left.\vphantom{ u;5}\right)$du =   0

for the mean and  

$\displaystyle \int_{{-\infty }}^{{\infty }}$u2 $\displaystyle \limfunc$TDen$\displaystyle \left(\vphantom{ u;5}\right.$u;5$\displaystyle \left.\vphantom{ u;5}\right)$du =  $\displaystyle {\frac{{25}}{{3}}}$$\displaystyle \sqrt{{5}}$

for the variance. (When the parameters are symbolic, there are problems in carrying out computations of the integrals that give mean and variance. You should have no difficulty when you specify numerical parameters.) BITMAPSETProbSolvHint0.1807in0.2188in0inq4

5.
Following is a sample solution.

6.
The probability of getting a 4 on a single cast is ${\frac{{1}}{{6}}}$, so the probability of getting a different result is ${\frac{{5}}{{6}}}$. The probability of casting the die five times without getting a 4 is $\left(\vphantom{ \frac{5}{6}}\right.$${\frac{{5}}{{6}}}$$\left.\vphantom{ \frac{5}{6}}\right)^{{5}}_{}$ =  .40188.BITMAPSETProbSolvHint0.1807in0.2188in0inq6

7.
With 600 calls on average during rush hour, the average number of calls per minute is 10. The probability that the number of connections in a given minute is less than or equal to 20 is the sum $\sum_{{k=0}}^{{20}}$$\limfunc$PoissonDen(k, 10) = $\sum_{{k=0}}^{{20}}$${\frac{{10^{k}e^{-10}%
}}{{k!}}}$ =  .99841. Thus, the probability that the board will be overtaxed is - .99841 =  .00159.BITMAPSETProbSolvHint0.1807in0.2188in0inq7

8.
x2≤4 when -2≤x≤2. So BITMAPSETProbSolvHint0.1807in0.2188in0inq8
Pr$\displaystyle \left(\vphantom{ x^{2}\leq 4}\right.$x2≤4$\displaystyle \left.\vphantom{ x^{2}\leq 4}\right)$ = Pr(x≤2) - Pr(x≤ - 2)  
  = $\displaystyle \func$NormalDist(2;1, 1) - $\displaystyle \func$NormalDist(- 2;1, 1)  
  = .84  



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