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26 August, 15:18

The time taken to prepare the envelopes to mail a weekly report to all executives in a company has a normal distribution, with a mean of 35 minutes and a standard deviation of 2 minutes. On 95% of occasions the mailing preparation takes less than a) 38.29 minutes b) 31.71 minutes c) 35.25 minutes d) 34.75 minutes

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  1. 26 August, 18:27
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    Mailing preparation takes 38.29 min max time to prepare the mails.

    Step-by-step explanation:

    Given:

    Mean:35 min

    standard deviation:2 min

    and 95% confidence interval.

    To Find:

    In normal distribution mailing preparation time taken less than.

    i. eP (t
    Solution:

    Here t - time and x - required time

    mean time 35 min

    5 % will not have true mean value. with 95 % confidence.

    Question is asked as, preparation takes less than time means what is max time that preparation will take to prepare mails.

    No mail take more time than that time.

    by Z-score or by confidence interval is

    Z = (X-mean) / standard deviation.

    Z=1.96 at 95 % confidence interval.

    1.96 = (X-35) / 2

    3.92 = (x-35)

    X=38.29 min

    or

    Confidence interval = 35±Z*standard deviation

    =35±1.96*2

    =35±3.92

    =38.29 or 31.71 min

    But we require the max time i. e 38.29 min

    And by observation we can also conclude the max time from options as 38.29 min.
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