# Normal and Binomial Distributions: Random Variables

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1) A videotape store has an average weekly gross of $1,158 with a standard deviation of $120. Assuming this to be a normally distributed random variable, calculate the following:

a. Probability the weekly gross will exceed $1,261.

b. Proportion of weeks the weekly gross is less than $1,080.

c. Probability the weekly gross is between $1,050 and $1,350.

d. A random sample of 12 weeks of historical, actual weekly gross sales was chosen. Find the probability that no more than 4 of those weeks exhibited weekly gross sales in excess of $1,261.

2) Bank robbers brandish firearms to threaten their victims in 70 percent of the incidents. An announcement that five bank robberies are taking place is being broadcast.

a. How many robberies would you expect to be happening from this group of five in which a firearm is being used?

b. What is the probability that a firearm is being used in at least two of the robberies?

3) A company sells toothpaste in a tube advertised to contain 8.00 ounces. The tube filling process is set with a mean of 8.25 ounces. In this continuous production process, the amount of toothpaste put in a tube is normally distributed with a mean of 8.25 ounces and a standard deviation of 0.09 ounces. If the actual capacity of the tubes used is 8.45 ounces, what proportion of the tubes will be filled beyond capacity?

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##### Solution Summary

This solution looks at applications with normally distributed random variables and binomial distributions: videotape store, bank robbery, and toothpaste.

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Problem 1

A.

The z-score will be (1,261 - 1,158)/120 = 0.86.

Probability (z > 0.86) = 0.1949.

B.

The z-score will be (1,080 - 1,158)/120 = -0.65.

Probability (z < -0.65) = 0.2578.

26% of the weeks or 26 out of 100 weeks will have a weekly gross of less than $1,080.

C.

The two z scores will be: z1 = (1,050 - 1,158)/120 = -0.90 and z2 = (1,350 - 1,158)/120 = 1.60.

Probability (-0.90 < z < 1.60) = 0.9452 - 0.1841 = ...

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