Standard deviation is a useful concept in performance management. Let us say that a director in a local fire department wants to know any variation between the performance of this year and that of the last year. He draws a sample of 10 response times of this year ( in minutes): 3.0, 12.0, 7.0, 4.0, 4.0, 6.0, 3.0, 9.0, 11.0, and 15.0, comparing them with a sample of 10 response times last year ( in minutes): 8.0, 7.0, 8.0, 6.0, 6.0, 9.0, 7.0, 9.0, 8.0, and 6.0.

Required:
a. Does he see a performance variation by the mean?
b. Does he see a performance variation by the standard deviation? If he does, is it performance improvement or deterioration from the last year? Why?

Step-by-step explanation:

Given the data :

Response time this year :

3.0, 12.0, 7.0, 4.0, 4.0, 6.0, 3.0, 9.0, 11.0, 15.0

Response time last year :

8.0, 7.0, 8.0, 6.0, 6.0, 9.0, 7.0, 9.0, 8.0, 6.0

Let's calculate the mean and standard deviation of each data sample :

Mean(m) = Σ(X) / n

n = sample size = 4

This year:

ΣX = 74 ; n = 10

Mean = 74 /10 = 7. 4

Standard deviation(s) = √(Σ(x - m)²/n - 1)

s = 4.195

Using a calculator to save computation time :Standard deviation = 4.195

LAST YEAR :

ΣX = 74 ; n = 10

Mean = 74 /10 = 7. 4

Standard deviation(s) = √(Σ(x - m)²/n - 1)

Using a calculator to save computation time :Standard deviation = 1.17

a. Does he see a performance variation by the mean?

From the mean value obtained for the two years, both are the same, hence, the value does not signify a variation in performance.

b. Does he see a performance variation by the standard deviation? If he does, is it performance improvement or deterioration from the last year? Why?

Yes, the value of standard deviation are different with the current year having a value of 4.195 and the previous year, a value of 1.17

The variation shows there is a deterioration on this year's performance from last year due to the larger value of standard deviation obtained ; 4.195 > 1.17

c is the answer

step-by-step explanation: