Example 7.3 from Statistics, 7.1 The Central Limit Theorem for Sample Means (Averages)

The Central Limit Theorem for Sample Means (Averages): example 7.3

In a recent study reported Oct. 29, 2012, the mean age of tablet users is 34 years. Suppose the standard deviation is 15 years. Take a sample of size n = 100.

What are the mean and standard deviation for the sample mean ages of tablet users?
What does the distribution look like?
Find the probability that the sample mean age is more than 30 years (the reported mean age of tablet users in this particular study).
Find the 95^th percentile for the sample mean age (to one decimal place).

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. Because the sample mean tends to target the population mean, we have μ_χ = μ = 34. The sample standard deviation is given by $σ_{χ}$ = $\frac{σ}{\sqrt{n}}$ = $\frac{15}{\sqrt{100}}$ = $\frac{15}{10}$ = 1.5.
2. The central limit theorem states that for large sample sizes (n), the sampling distribution will be approximately normal.
3. The probability that the sample mean age is more than 30 is given by P(Χ > 30) = normalcdf(30,E99,34,1.5) = 0.9962.
4. Let k = the 95^th percentile.
  k = invNorm $(0. 95,34, \frac{15}{\sqrt{100}})$ = 36.5

How did it go?

Solution

Solution