Standard Deviation and Expected Value of Random Variables

A hat contains 5 slips of paper, numbered 1 through 5, with a different number on each slip of paper. Another hat contains 4 slips of paper, numbered 20 through 23, with a different number on each slip of paper. One number is randomly drawn from each hat. What is the expected value of the sum of the numbers on the two selected pieces of paper?

• 12.25

• 12.5

• 18.5

• 24.5

• 25

The U.S. Postal Service recently released data on the number of days it takes for a first-class letter to reach its destination, as shown in the table above.

• μ = 1.9; σ = 0.75

• μ = 1.9; σ = 0.87

• μ = 2; σ = 1

• μ = 2; σ = 1.5

• μ = 2.5; σ = 1

The U.S. Postal Service recently released data on the number of days it takes for a first-class letter to reach its destination, as shown in the table above.

Two first-class letters are randomly selected from all letters mailed on a given day. Let Z be the sum of the number of days it takes for the two letters to reach their respective destinations.

What are the mean and standard deviation of the random variable Z?

• μ = 1.9; σ = 1.22

• μ = 1.9; σ = 1.5

• μ = 3.8; σ = 0.75

• μ = 3.8; σ = 1.22

• μ = 3.8; σ = 1.5

A is a random variable whose mean is 71.2 and whose standard deviation is 1.1. B is a random variable whose mean is 23.7 and whose standard deviation is 0.25.

If A and B are independent, what is the variance of A - B ?

• 0.0625

• 1.21

• 1.147

• 1.2725

• 1.35

A small fry weighs 3 ounces with a standard deviation of 0.2 ounces and a cheeseburger weighs 5 ounces with a standard deviation of 0.4 ounces. Jimmy makes a heart attack bet with one of his friends that he can eat 4 small fries and 4 cheeseburgers in 10 minutes.

Not only do we want to know if Jimmy can accomplish this heart attack goal, but what is the total mean and standard deviation in ounces for the cheeseburgers and small fries that Jimmy is about to devour?

• Mean = 8 ounces

  Standard Deviation = 0.8 ounces

• Mean = 8 ounces

  Standard Deviation = 0.89 ounces

• Mean = 32 ounces

  Standard Deviation = 2.4 ounces

• Mean = 32 ounces

  Standard Deviation = 0.8 ounces

• Mean = 32 ounces

  Standard Deviation = 0.89 ounces

Marlene works two jobs to help make ends meet. As a waitress, her weekly earnings (including tips) are $590 approximately normally distributed with a mean of and a standard deviation of $45.

As a dog walker, her weekly earnings depend upon how many dog walks she provides that week. The number of dog walks she provides each week is a random variable (D) whose probability distribution is shown above.

If Marlene $20 charges per dog walk, on average, how much can she expect to make each week from both of her two jobs combined?

• $7.01

• $140.01

• $590.00

• $730.20

• Since Marlene's earnings as a dog walker are not approximately normally distributed, we cannot find the mean sum of her earnings from both jobs combined.

Suppose is a random variable X where the mean is 14.2 and the standard deviation is 3.2.

Let y = 5.1x + 8 What is the mean and variance of y?

• Mean: 80.42 Variance: 274.34

• Mean:80.42 Variance: 266.34

• Mean: 80.42 Variance: 16.32

• Mean: 72.42 Variance: 274.34

• Mean: 72.42 Variance: 266.34

Homer Simpson has yet another new job. He is now the personal chauffeur for a bunch of characters. These characters always travel together to places like the CartoonCon Convention, and Homer always drives them in a van that weighs 4050 pounds.

Of course, Homer's weight and the weights of the other characters go up and down a bit each day depending on how much they have been enjoying their food. The table above gives Homer's and each of the other character's mean weights and the standard deviation of their weights (in pounds).

• Mean = 582.7 pounds

  Standard Deviation = 1.6 pounds

• Mean = 582.7 pounds

  Standard Deviation = 7.62 pounds

• Mean = 4632.7 pounds

  Standard Deviation =1.6 pounds

• Mean = 4632.7 pounds

  Standard Deviation = 7.62 pounds

• We cannot find the standard deviation of the total weight since we are not given the standard deviation of the weight of the van.