Example 7.7 from Statistics, 7.2 The Central Limit Theorem for Sums (Optional)

The Central Limit Theorem for Sums (Optional): example 7.7

The mean number of minutes for app engagement by a tablet user is 8.2 minutes. Suppose the standard deviation is one minute. Take a sample size of 70.

What are the mean and standard deviation for the sums?
Find the 95^th percentile for the sum of the sample. Interpret this value in a complete sentence.
Find the probability that the sum of the sample is at least 10 hours.

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

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Solution
1. μ_Σx = nμ_x = 70(8.2) = 574 minutes and σ_Σx = $(\sqrt{n}) (σ_{x})$ = $(\sqrt{70})$ (1) = 8.37 minutes
2. Let k = the 95^th percentile.
  k = invNorm (0.95,(70)(8.2), $(\sqrt{70})$ (1)) = 587.76 minutes
  Ninety-five percent of the app engagement times are at most 587.76 minutes.
3. 10 hours = 600 minutes
  P(Σx ≥ 600) = normalcdf(600,E99,(70)(8.2), $(\sqrt{70})$ (1)) = 0.0009

How did it go?

Solution

Solution