Question

The data below indicates the time (in seconds) that it takes 25 separate employees to complete a certain task.  

hw1_q5a

a) What is the mean of the data in seconds? Give your answer to three decimal places.  

b)What is the median of the data in seconds?  

c)What is the variance of the data in seconds? Give your answer to three decimal places.  

d)What is the standard deviation of the data in seconds? Give your answer to three decimal places.  

e)What is the mean of the data in minutes? Give your answer to three decimal places.  

f)What is the median of the data in minutes? Give your answer to three decimal places.  

g)What is the variance of the data in minutes? Give your answer to three decimal places.  

h)What is the standard deviation of the data in minutes? Give your answer to three decimal places.

seconds: hw1_q5a

seconds 289 221 226 255 219 218 261 189 233 269 226 270 255 233 213 197 210 229 240 292 228 341 183 228 293

Answer 1

a)

Mean of the data is,

= (289+221+226+255+219+218+261+189+233+269+226+270+255+233+213+197+210+229+240+292+228+341+183+228+293) / 25 = 240.72 seconds

b)

Data is ascending order is,

183 189 197 210 213 218 219 221 226 226 228 228 229 233 233 240 255 255 261 269 270 289 292 293 341

For odd sample size, median is the (n+1)/2 th element = (25 + 1)/2 = 13th element

Thus, the median = 229 seconds

c)

Variance of the data is,

= [(289 -240.72)² + (221 -240.72)² + (226 -240.72)² + (255 -240.72)² + (219 -240.72)² + (218 -240.72)² + (261 -240.72)² + (189 -240.72)² + (233 -240.72)² + (269 -240.72)² + (226 -240.72)² + (270 -240.72)² + (255 -240.72)² + (233 -240.72)² + (213 -240.72)² + (197 -240.72)² + (210 -240.72)² + (229 -240.72)² + (240 -240.72)² + (292 -240.72)² + (228 -240.72)² + (341 -240.72)² + (183 -240.72)² + (228 -240.72)² + (293 -240.72)² ] / 24 = 1343.627 seconds²

d)

Standard deviation of the data = $\sqrt{1343.627}$ = 36.656 seconds

e)

Let S and M be the data in seconds and minutes. Then M = S/60

E[M] = E[S / 60] = E[S] / 60

Thus, mean of the data in minutes = 240.72 / 60 = 4.012 minutes

f)

Median of the data in minutes =  229/60 = 3.817 minutes

g)

Var[M] = Var[S / 60] = Var[S] / 60² = Var[S] / 3600

Thus, variance of the data in minutes = 1343.627 / 3600 = 0.373 minutes

h)

Standard deviation of the data in minutes = $\sqrt{0.373}$ = 0.611 minutes