Help with a statistics test question. Confidence intervals and sample size

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The life of a freeze breeze electric fan is normally distributed with a population mean= 48 months and a population standard deviation = 14 months. The manufacturer will replace any fan that wears out during that guarantee period. (a) what percentage of fans will have to be replaced if the guarantee period is 36 months? Answer to 2 decimal places. (b) what should the guarantee period be if the manufacturer does not want to replace more than 5% of the fans? Answer to the nearest month.

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@Krommydasmary,

Start with the calculator on this site: http://stattrek.com/online-calculator/normal.aspx

Enter a mean of 48
Enter a standard deviation of 14
Now enter 36 as the Normal Random Variable and press calculate.
19.57%

Enter values less than 36 in the Normal Random Variable field and press calculate until you get less than 0.05 (5%) as a result. (You have to delete the previous answer for each guess.)
25 gives you 5.01%, so if you want no more that 5%, you should go with 24 months.

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@Krommydasmary,

If you want the theory for the test:

z = (x - mean) / sigma
z = (36 - 48)/14 = -0.857
Look up the cumulative normal distribution for that z score and you get 0.1956

For the second part, look up the z score for a cumulative normal distribution value of 0.05. It is -1.645, so that is z.

-1.645 = (x - 48)/14
x = 24.97 months (round down to ensure that the company pays no more that 5%)

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Thanks so much! I appreciate it:)

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Help with a statistics test question. Confidence intervals and sample size

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