Question

The probability distribution for the random variable x follows. X F(x) 20 . 20 25 .15 30 .25 35 .40 a. Is this probability distribution valid? Explain. b. What is the probability that x = 30? c. What is the probability that x is less than or equal to 25? d. What is the probability that x is greater than 30? E. Compute the expected value, variance, and standard deviation of x. Employee retention is a major concern for many companies. A survey of Americans asked how long they have worked for their current employer (Bureau of Labor Statistics website). Consider the following example of sample data of 2000 college graduates who graduated five years ago.

Answer 1

a) Since the sum of the probabilities is 1, the probability distribution is valid.

b) The probability that x = 30 is 0.25.

c) The probability that x is less than or equal to 25 is 0.35.

d) The probability that x is greater than 30 is 0.40.

e) The expected value, variance, and standard deviation of (x) are 28, 20, and 4.47, respectively.

a. Validity of the probability distribution:

The sum of the probabilities = 1.00 [0.20 + 0.15 + 0.25 + 0.40]

A probability distribution is valid if the sum of all probabilities is equal to 1, each probability is between 0 and 1, and each probability is between 0 and 1. So, this is a valid probability distribution.

X F(x)

20 0.20

25 0.15

30 0.25

35 0.40

Total = 1 (0.2 + 0.15 + 0.25 + 0.4)

b. The probability that x = 30:

From the table, the probability that x = 30 is given as 0.25.

c. The probability that x is less than or equal to 25:

The probability that x is less than or equal to 25 is the sum of the probabilities for x = 20 and x = 25, which is 0.20 + 0.15

= 0.35

d. The probability that x is greater than 30:

The probability that x is greater than 30 is the probability for x = 35, which is 0.40.

e. To compute the expected value, variance, and standard deviation of (x), we can use the following formulas:

Expected Value
`((E(x))): [E(x) = \sum x \cdot P(x)] [E(x) = (20 \cdot 0.20) + (25 \cdot 0.15) + (30 \cdot 0.25) + (35 \cdot 0.40)] [E(x) = 28]`

Variance ((Var(x))):
`[Var(x) = \sum (x - E(x))^2 \cdot P(x)] [Var(x) = (20 - 28)^2 \cdot 0.20 + (25 - 28)^2 \cdot 0.15 + (30 - 28)^2 \cdot 0.25 + (35 - 28)^2 \cdot 0.40] [Var(x) = 20]`

Standard Deviation
`((\sigma)): [\sigma = √(Var(x))] [\sigma = √(20)] [\sigma \approx 4.47]`

Thus, the probability distribution is valid, and the probability that (x = 30) is 0.25. The probability that (x) is less than or equal to 25 is 0.35, and the probability that (x) is greater than 30 is 0.40. The expected value of (x) is 28, the variance is 20, and the standard deviation is approximately 4.47.