Question

Complete parts (a) and (b) using the probability distribution below.

The number of overtime hours worked in one week per employee

Overtime hours
0 1 2 3 4 5 6

Probability
0.026 0.072 0.154 0.303 0.215 0.164 0.066

(a) Find the mean, variance, and standard deviation of the probability distribution.

Find the mean of the probability distribution.

μ = 3.4

σ2 = 2.0

σ = 1.4
Choose the correct answer below.

A.
An employee works an average of 2.0 overtime hours per week with a standard deviation of approximately 1.4 hours.

B.
An employee works an average 3.4 of overtime hours per week with a standard deviation of approximately 4 hours.

C.
An employee works an average of 1.4 overtime hours per week with a standard deviation of approximately 3.4 hours.

D.
An employee works an average of 3.4 overtime hours per week with a standard deviation of approximately 1.4 hours.

Answer 1

μ = (0×0.026) + (1×0.072) +(2×0.152) + (3×0.303) + (4×0.215) + (5×0.164) + (6×0.066) 
μ = 0 + 0.072 + 0.304 + 0.909 + 0.86 + 0.82 + 0.396
μ = 3.361 ≈ 3.4

We need the value of ∑X² to work out the variance
∑X² = (0²×0.026) + (1²×0.072) + (2²×0.152) + (3²×0.303) + (4²×0.215) + (5²×0.164) + (6²×0.066)
∑X² = 0+0.072+0.608+2.727+3.44+4.1+2.376
∑X² = 13.323

Variance = ∑X² - μ²
Variance  = 13.323 - (3.4)² = 1.763 ≈ 2

Standard Deviation = √Variance = √1.8 = 1.3416... ≈ 1.4

The correct answer related to the value of mean and standard deviation is the option D

An employee works an average of 3.4 overtime hours per week with a standard deviation of approximately 1.4 hours.