5. Answer the following in details and steps. Thanks.a. The length of time it takes to fill an order at a ABC sandwich shop is normally distributed with a mean of 6 minutes and a standard deviation of 2 minute. What is the probability that the average waiting time for a random sample of 100 customers is between 5.6 and 6 minutes?  b. The length of time it takes to fill an order at a DEF sandwich shop is normally distributed with a mean of 5 minutes and a standard deviation of 2 minute. Suppose the probability that the average waiting time for a random sample of customers between 4.75 and 5 minutes is 0.3413. What is the number of customers in the sample? c. The length of time it takes to fill an order at a XYZ sandwich shop is normally distributed with a mean of 7 minutes. Suppose the probability that the average waiting time for a random sample of 36 customers between 6.8 and 7.2 minutes is 0.383. What is the standard deviation of the waiting time?

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Asked Dec 16, 2019
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5. Answer the following in details and steps. Thanks.

a. The length of time it takes to fill an order at a ABC sandwich shop is normally distributed with a mean of 6 minutes and a standard deviation of 2 minute. What is the probability that the average waiting time for a random sample of 100 customers is between 5.6 and 6 minutes?

 

 

b. The length of time it takes to fill an order at a DEF sandwich shop is normally distributed with a mean of 5 minutes and a standard deviation of 2 minute. Suppose the probability that the average waiting time for a random sample of customers between 4.75 and 5 minutes is 0.3413. What is the number of customers in the sample?

 

c. The length of time it takes to fill an order at a XYZ sandwich shop is normally distributed with a mean of 7 minutes. Suppose the probability that the average waiting time for a random sample of 36 customers between 6.8 and 7.2 minutes is 0.383. What is the standard deviation of the waiting time?

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P(5.6 < X < 6) = P(X < 6) – P(X < 5.6) Changing into standard normal variate 5.6 – 6 5.6 — и 6 - u 6 – 6 = P z < Vn vn yn V100 V100 = P(z < 0) – P(z<-2) = NORM. S. DIST(0, TRUE) (P(z < 0) = 0.5 From excel (P(z <-2) = 0.0228 From excel = NORM. S. DIST(-2, TRUE) 0.5 – 0.0228 = 0.4772

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