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25 July, 05:04

A popular retail store knows that the purchase amounts by its customers is a random variable that follows a normal distribution with a mean of $30 and a standard deviation of $9. what is the probability that a randomly selected customer will spend between $20 and $35 at this store? place your answer, rounded to 4 decimal places, in the blank. for example, 0.3456 would be a legitimate entry.

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  1. 25 July, 06:26
    0
    With continuity correction. 20-35 becomes 19.5 to 35.5

    z1 = (19.5-30) / 9 = - 1.167

    z2 = (35.5-30) / 9 = 0.61

    P (-1.167 < z < 0.61) = 0.6075
  2. 25 July, 07:33
    0
    We are going to use a continuity correction factor. When you use a normal distribution to get the estimated binomial distribution, you're going to have to use a continuity correction factor. So with this, the data will become from 20 - 35 it will now become 19.5 to 35.5.

    So we are going to use the z table for this problem. So if we are looking for the z value, the formula would be: x - μ / σ

    get the value of the first z value:

    z1 = (19.5 - 30) / 9 = - 1.167

    get the value of the second z value:

    z2 = (35.5 - 30) / 9 = 0.61

    So getting the probability of both:

    P (-1.167 < z < 0.61) = P (Z ≤ b) - P (Z ≤ a) = 0.6075
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