The average hourly wage of fast-food workers employed by a nationwide chain is $5.55. The standard deviation is $1.15. If a sample of 50 workers is selected, find the probability that the mean of the sample will be between $5.25 and $5.90.
Since this problem deals with a distribution of averages, the Cental Limit Theorem applies. Thus, the distribution is normal with standard deviation of . The TI83/84 has a function that preforms the required calculation. That function is
From the statement of the problem, we have = 5.55, = 1.15, n = 50, = 5.25, and = 5.90.
From the home screen, press 2nd Vars[DISTR] and then press 2 to select 2:normalcdf(. Press ENTER. Type . The results are shown below.
Thus, the probability that the mean of the sample will be between $5.25 and $5.90 is about 95%.
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