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25 January, 00:29

The time taken by a randomly selected applicant for a mortgage to fill out a certain form has a normal distribution with mean value 10 min and standard deviation 2 min. if five individuals fill out a form on one day and six on another, what is the probability that the sample average amount of time taken on each day is at most 11 min?

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  1. 25 January, 00:36
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    To solve for the probability proportion, we make use of the z statistic. The procedure to do is to calculate for the z value and using the standard probability tables, we can look up for the p value. The formula for z score is:

    z = (x - μ) / (σ / sqrt (n))

    where,

    x = sample score = 11

    μ = sample mean = 10

    σ = standard deviation = 2

    n = sample size

    Calculating for the z and p value when n = 5:

    z = (11 - 10) / (2 / sqrt (5))

    z = 1.12

    Using the tables, p (5) = 0.8686

    Calculating for the z and p value when n = 6:

    z = (11 - 10) / (2 / sqrt (6))

    z = 1.22

    Using the tables, p (6) = 0.8888

    If both days should be occuring, therefore the total probability that each day is at most 11 min is:

    p total = p (5) * p (6)

    p total = 0.8686 * 0.8888

    p total = 0.772
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