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Today, 12:31

A random sample of 12 graduates of a certain secretarial school typed an average of 79.3 words per minute with a standard deviation of 7.8 words per minute. assuming a normal distribution for the number of words typed per minute, find a 95% confidence interval for the average number of words typed by all graduates of this school.

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  1. Today, 12:52
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    We will use the following formula to work out the confidence interval

    Upper limit = μ + z * (σ/√n)

    Lower limit = μ - z * (σ/√n)

    We have

    μ = 79.3

    σ = 7.8

    n = 12

    z * is the z-score for 95% confidence level = 1.96

    Substitute these into the formula, we have

    Upper limit = 79.3 + 1.96 (7.8/√12) = 83.7

    Lower limit = 79.3 - 1.96) 7.8/√12) = 74.9
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