Question

In: Statistics and Probability

Suppose a restaurant determines that their daily profits are approximately normally distributed with a mean of...

Suppose a restaurant determines that their daily profits are approximately normally distributed with a mean of μ = $1500 and a standard deviation of σ = $1050.

  1. What is the probability that the restaurant makes a profit each day? In other words, what is the probability that the restaurant’s daily profit is at least one cent?

  2. Assuming that the standard deviation stays the same, what daily mean would be re- quired for there to be a 99% chance that the restaurant’s daily profits are at least one cent? [Hint: Find the negative z-score that corresponds to a right tail probability of .99. This will allow you to determine the mean.]

Solutions

Expert Solution

We are given that a restaurant's daily profits are normally distributed with mean and standard deviation of and are asked to find the probability that the restaurant's daily profits are atleast one cent.

So, we are asked to find .

We know that =1-P(X<.01) which can be calculated by using the z-score formula.

So now, P(X<0.01) will be the value obtained under the z-score.

From the standard normal tables, the value under the z-score of -1.43 is 0.0764.

Hence

Therefore the probability that the restaurant makes a daily profit of 1 cent is 92.36%.

We are now asked to find the mean for which the chance of daily profits are atleat one cent is 99% given that the standard deviation remains the same.

Since we are finding a negative z-score to obtain the mean, our probability or the chance of daily profits are atleat one cent becomes 1-0.99=0.01.

Looking at the standard normal tables for the z-score corresponding to the value 0.01, we find two values closer to 0.01 neamely. 0.0102 and 0.0099 corresponding to the z-scores -2.32 and -2.33 respectively.

We find the approximate z-score corresponding to 0.01 by interpolation as follows:

this represents the proportional distance from top.

Hence -2.3267 is the required z-score

Now using the z-score formula we can find the mean to be a 99% chance that the restaurant's daily profits are atleast one cent

Hence the required mean is $2443.045. to be a chance that the restaurant's daily profits are atleast one cent.


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