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11 July, 06:01

Suppose that a random sample of size 36 is to be selected from a population with mean 50 and standard deviation 7. What is the approximate probability that X will be within. 5 of the population mean

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  1. 11 July, 09:27
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    Step-by-step explanation:

    Let us assume that x is normally distributed. The sample size is greater than 30. Since the the population mean and population standard deviation are known, we would apply the formula,

    z = (x - µ) / (σ/√n)

    Where

    x = sample mean

    µ = population mean

    σ = standard deviation

    n = number of samples

    From the information given,

    µ = 50

    σ = 7

    n = 36

    If x is within 0.5 of the population mean, it means that x is between (50 - 0.5) and (50 + 0.5)

    the probability is expressed as

    P (49.5 ≤ x ≤ 50.5)

    For x = 49.5

    z = (49.5 - 50) / (7/√36) = - 0.43

    Looking at the normal distribution table, the probability corresponding to the z score is 0.334

    For x = 49.5

    z = (50.5 - 50) / (7/√36) = 0.43

    Looking at the normal distribution table, the probability corresponding to the z score is 0.666

    Therefore,

    P (49.5 ≤ x ≤ 50.5) = 0.666 - 0.334 = 0.332
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