7.3 Using the central limit theorem  (Page 2/8)

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 $(.90,75\cdot 3,\sqrt{75}\cdot 1.15)=237.8$

Part a.

$P(\overline{x}> 20)=0.7919$ 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.

Part b.

P(X>20) = e^(–(1/22)*20) or e^(–.04545*20) = 0.4029

Part c. explain why the probabilities in (a) and (b) are different.

• $P(x> 20)=0.4029$ but $P(\overline{x}> 20)=0.7919$
• The probabilities are not equal because we use different distributions to calculate the probability for individuals and for means.
• When asked to find the probability of an individual value, use the stated distribution of its random variable; do not use the CLT. Use the CLT with the normal distribution when you are being asked to find the probability for an mean.

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.