Answer:
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