Question

Overtime hours Probability 0.045 0.070 0.164 (a) Find the mean, variance, and standard deviation of the probability distribution. Find the mean of the probability distribution. Round to one decimal place as needed.) Find the variance of the probability distribution. ơ2-Q (Round to one decimal place as needed) Find the standard deviation of the probability distribution. -(Round to one decimal place as needed ) (b) Interpret the results in the context of the real-life situation O A. An employee works an average of 3.3 overtime hours per week with a standard deviation of approximately 1.5 hours. OB. An employee works an average 33 of overtime hours per week with a standard deviation of approximately 9 hours Click to select your answer(s) O Type here to search

Answer 1

Answer)

A)

Mean is given by sum of x.p(x)

= 0*0.045 + 1*0.070 +2*0.164 + 3*0.269 + 4*0.229 + 5*0.139 + 6*0.084

= 3.32 is the required mean

or 3.3 hours approx

Steps to calculate variance

First subtract mean hours from each and every observation

0-3.32, 1-3.32, 2-3.32, 3-3.32, 4-3.32, 5-3.32, 6-3.32

= -3.32, -2.32, -1.32, -0.32, 0.68, 1.68, 2.68

now we need to square them all

11.0224, 5.3824, 1.7424, 0.1024, 0.4624, 2.8224, 7.1824

Now variance is given by the sum of (x-mean)² * P(x)

Therefore, required variance is

11.0224*0.045 + 5.3824*0.070 + 1.7424*0.164 +0.1024*0.269 + 0.4624*0.229 + 2.8224*0.139 + 7.1824*0.084

= 2.2876

or 2.3(rounded to one decimal place)

Standard deviation is square root of variance

square root of 2.3

= 1.5 (rounded to 1 decimal place)

Therefore

required mean = 3.3

Variance is = 2.3

Standard deviation is 1.5

B)

By above calculation, option A is correct.