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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 ( Σx k ) = 0.90 .

k = 237.8

Normal distribution curve of sum x with k on the x-axis. Vertical upward line extends from k to the curve. The probability area under the curve from the beginning of the curve to k is equal to 0.90.

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 3 , 75 1.15 ) = 237.8

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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 ~ Exp ( 1 22 ) From Chapter 5, we know that μ = 22 and σ = 22 .

Let X = the mean excess time used by a sample of n = 80 customers who exceed their contracted time allowance.

X ~ N ( 22 , 22 80 ) by the CLT for Sample Means

    Using the clt to find probability:

  • Find the probability that the mean excess time used by the 80 customers in the sample is longer than 20 minutes. This is asking us to find P ( x 20 ) Draw the graph.
  • Suppose that one customer who exceeds the time limit for his cell phone contract is randomly selected. Find the probability that this individual customer's excess time is longer than 20 minutes. This is asking us to find P ( x 20 )
  • Explain why the probabilities in (a) and (b) are different.

Part a.

Find: P ( x 20 )

P ( x 20 ) = 0.7919 using normalcdf ( 20 , 1E99 , 22 , 22 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.

Normal distribution curve with values of 20 and 22 on the x-axis. Vertical upward line extends from point 20 to curve. The probability area begins from point 20 to the end of the curve.

1E99 = 10 99 and -1E99 = - 10 99 . Press the EE key for E. Or just use 10^99 instead of 1E99.

Part b.

Find P(x>20) . Remember to use the exponential distribution for an individual: X~Exp(1/22) .

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 ( 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.
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Using the clt to find percentiles:

Find the 95th percentile for the sample mean excess time for samples of 80 customers who exceed their basic contract time allowances. Draw a graph.

Let k = the 95th percentile. Find k where P ( x k ) = 0.95

k = 26.0 using invNorm ( .95 , 22 , 22 80 ) = 26.0

Normal distribution curve with value of k on x-axis. Vertical upward line extends from k to curve. Probability area from the beginning of the curve to point k is equal to 0.95.

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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Source:  OpenStax, Collaborative statistics. OpenStax CNX. Jul 03, 2012 Download for free at http://cnx.org/content/col10522/1.40
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