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Find the 90th percentile for the total of 75 stress scores. Draw a graph.
Let $k$ = the 90th percentile.
Find $k$ where $P(\mathrm{\Sigma x}< k)=0.90$ .
$k=237.8$
The 90th percentile for the sum of 75 scores is about 237.8. This tells us that 90% of all the sums of 75 scores are no more than 237.8 and 10% are no less than 237.8.
invNorm
$(.90,75\cdot 3,\sqrt{75}\cdot 1.15)=237.8$
Suppose that a market research analyst for a cell phone company conducts a study of their customers who exceed the time allowance included on their basic cell phone contract; the analyst finds that for those people who exceed the time included in their basic contract, the excess time used follows an exponential distribution with a mean of 22 minutes.
Consider a random sample of 80 customers who exceed the time allowance included in their basic cell phone contract.
Let $X$ = the excess time used by one INDIVIDUAL cell phone customer who exceeds his contracted time allowance.
$X$ ~ $\mathrm{Exp}\left(\frac{1}{22}\right)$ From Chapter 5, we know that $\mu =22$ and $\sigma =22$ .
Let $\overline{X}$ = the mean excess time used by a sample of $n=80$ customers who exceed their contracted time allowance.
$\overline{X}$ ~ $N(22,\frac{22}{\sqrt{80}})$ by the CLT for Sample Means
$P(\overline{x}> 20)=0.7919$ using
normalcdf
$(20,\mathrm{1E99},22,\frac{22}{\sqrt{80}})$
The probability is 0.7919 that the mean excess time used is morethan 20 minutes, for a sample of 80 customers who exceed their contracted time allowance.
EE
key for E. Or just use 10^99 instead of 1E99.
P(X>20) = e^(–(1/22)*20) or e^(–.04545*20) = 0.4029
Let $k$ = the 95th percentile. Find $k$ where $P(\overline{x}< k)=0.95$
$k=26.0$ using
invNorm
$(.95,22,\frac{22}{\sqrt{80}})=26.0$
The 95th percentile for the sample mean excess time used is about 26.0 minutes for random samples of 80 customers who exceed their contractual allowed time.
95% of such samples would have means under 26 minutes; only 5% of such samples would have means above 26 minutes.
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