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Exercises Probability and the bell curve

Example 7.4 from Statistics, 7.1 The Central Limit Theorem for Sample Means (Averages)

The Central Limit Theorem for Sample Means (Averages): example 7.4

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 of 60.

What are the mean and standard deviation for the sample mean number of app engagement minutes by a tablet user?
What is the standard error of the mean?
Find the 90^th percentile for the sample mean time for app engagement for a tablet user. Interpret this value in a complete sentence.
Find the probability that the sample mean is between eight minutes and 8.5 minutes.

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.

1
Solution
1. $μ_{\bar{x}} = μ = 8.2 σ_{\bar{x}} = \frac{σ}{\sqrt{n}} = \frac{1}{\sqrt{60}} = 0.13$
2. This allows us to calculate the probability of sample means of a particular distance from the mean, in repeated samples of size 60.
3. Let k = the 90^th percentile.
  k = invNorm $(0. 90,8 .2, \frac{1}{\sqrt{60}})$ = 8.37. This values indicates that 90 percent of the average app engagement time for table users is less than 8.37 minutes.
4. P(8 < $\bar{x}$ < 8.5) = normalcdf $(8,8 .5,8 .2, \frac{1}{\sqrt{60}})$ = 0.9293

How did it go?

PreviousThe Central Limit Theorem for Sample Means (Averages): example 7.3 Next The Central Limit Theorem for Sums (Optional): example 7.5

From Statistics (High school edition) by OpenStax (Rice University), section 7.1, licensed under CC BY 4.0. Original at openstax.org. We split the solution into steps; the words, numbers and figures are the book's. OpenStax doesn't endorse this site.