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10 February, 09:57

A manufacturer produces 25-pound lifting weights. The lowest actual weight is 24 pounds, and the highest is 26 pounds. Each weight is equally likely so the distribution of weights is uniform. A sample of 100 weights is taken.

a. What is the distribution for the weights of one 25-pound lifting weight? What is the mean and standard deviation?

b. What is the distribution for the mean weight of 100 25-pound lifting weights?

c. Find the probability that the mean actual weight for the 100 weights is less than 24.9.

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  1. 10 February, 10:26
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    a) Mean = 25, S. d = 0.5774

    b) X~N (25, 0.05774)

    c) P (X < 24.9) = 0.0416

    Step-by-step explanation:

    Given:

    - Limits of a uniform distribution are:

    U (24, 26)

    - Sample size n = 100

    Find:

    a. What is the distribution for the weights of one 25-pound lifting weight? What is the mean and standard deviation?

    Solution

    The mean and standard deviation value is:

    E (X) = (24 + 26) / 2 = 25

    Var (X) = (25 - 24) / 3 = 1/3

    s. d (X) = sqrt (1/3) = 0.5774

    Find:

    b. What is the distribution for the mean weight of 100 25-pound lifting weights?

    Solution

    The random variable X follows a normal distribution:

    X~N (25, s. d/100)

    X~N (25, 0.05774)

    Find:

    c. Find the probability that the mean actual weight for the 100 weights is less than 24.9.

    Solution

    The probability of P (X < 24.9) is:

    P (X < 24.9) = P (Z < (24.9-25) / 0.05774)

    = P (Z < - 1.7319)

    = 0.0416
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