Question

The company’s cost of manufacturing a single 1-inch binder is $0.25. What is the expected profit (in dollars) that the company will make, per customer, for 1-inch binder sales during the promotion? The expected profit is 1.25
c) The mean and standard deviation of “Y = the number of 2-inch binders purchased by a single customer” is 2.74 and 1.25, respectively. Assume that the number of 1-inch and 2-inch binders purchased are independent random variables. What are the mean and standard deviation of the total number of binders purchased by a single customer?



d) The company’s cost of manufacturing a single 2-inch binder is $0.55. What is the total expected profit (in dollars) that the company will make, per customer, during the promotion?

Answer 1

It should be noted that the mean and the standard deviation of the data given in will be 4.26 and 1.1011 respectively.

How to solve the mean.

From the complete information, the mean will be calculated as:

= (0.02 × 1) + (0.09 × 2) + (0.12 × 3) + (0.15 × 4) + (0.62 × 5)

E(X) = 4.26

The standard deviation will be calculated thus:

E(X²) = (0.02 × 1²) + (0.09 × 2²) + (0.12 × 3²) + (0.15 × 4²) + (0.62 × 5²)

E(X²) = 19.36

The standard deviation will then be the difference between the square root of 19.36 and 4.26². This will be 1.1011.

The expected profit of the company will be:

= 0.75 × E(X)

= 0.75 × 4.26

= $3.20

The mean of the total number of binders purchased will be:

= 4.26 + 2.74 = 7

The standard deviation will be 1.6658.

Lastly, the total expected profit will be:

= 0.75E(X) + 1.45E(Y)

= 0.75(4.26) + 1.45(2.74)

= 3.195 + 3.973

= 71.68

Learn more about standard deviation on:

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