Question

In: Statistics and Probability

Anticipated consumer demand at a restaurant for free-range steaks next month can be modeled by a...

Anticipated consumer demand at a restaurant for free-range steaks next month can be modeled by a normal random variable with mean 1,500 pounds and standard deviation 110 pounds.

a. What is the probability that demand will exceed 1,300 pounds?

b. What is the probability that demand will be between 1,400 and 1,600 pounds?

c. The probability is 0.15 that demand will be more than how many pounds?

Expert Solution

Let "x" be the anticipated consumer demand at a restaurant

Refer Standard normal table/Z-table to find the probability or use excel formula "=NORM.S.DIST(-1.8182, TRUE)" to find the probability.

The probability that demand will exceed 1300 pounds is

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Refer Standard normal table/Z-table to find the probability or use excel formula "=NORM.S.DIST(0.9091, TRUE)" & "=NORM.S.DIST(-0.9091, TRUE)" to find the probability.

The probability that demand will be between 1,400 and 1,600 pounds is

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Refer Standard normal table/Z-table, Lookup for z-score corresponding to area 0.15 to the right of the normal curve or use excel formula "=NORM.S.INV(1-0.15)" to find the z-score.

The probability is 0.15 that demand will be more than pounds

orchestra answered 2 years ago

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