A Pizza delivery business in a large city has determined that the amount of time that a customer spends waiting for their delivery order is normally distributed with a mean of 26.5 minutes and a standard deviation of 3 minutes. What is the probability that for 100 randomly chosen customers the mean amount of time spent waiting for their delivery is greater than 27 minutes. Use the calculator to determine the probability Write the Calculator function

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jmonterrozar jmonterrozar · Answer 1 · 2020-04-12T02:27:36+02:00

Answer:

95.15%

Step-by-step explanation:

We have that the mean (m) is equal to 26.5, the standard deviation (sd) = 3 and the sample size (n) = 100

They ask us for P (x <27)

For this, the first thing is to calculate z, which is given by the following equation:

z = (x - m) / (sd / (n ^ 1/2))

We have all these values, replacing we have:

z = (27 - 26.5) / (3 / (100 ^ 1/2))

z = 1.66

With the normal distribution table (attached), we have that at that value, the probability is:

P (z <1.66) = 0.9515

The probability is 95.15%