Question

5. Question Details | Previous Answers My Notes Ask Your Teacher The data below indicate the time (in seconds) that it takes 25 separate employees to complete a certain task. hw1_q5.csv a) What is the mean of the data in seconds? Give your answer to three decimal places. 233.88 b)What is the median of the data in seconds? 217 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 Submit Answer Save Progress

seconds
260
224
221
255
236
215
193
285
244
284
222
185
182
197
253
191
255
231
229
168
152
221
244
239
246

Answer 1

For formula's used, please refer to the attached image.

for our case n=25, Sum of all observations=$\sum x_{i}$=5632

(A)Mean=$\sum x_{i}/n$=(5632/25)=225.280 second

(b) For this you need to arrange you observations either in ascending or descending order of sequence.

I have arranged them in ascending order. Data in ascending order is:

152,168,182,185,191,193,197,215,221,221,222,224,229,231,236,239,244,244,246,253,255,255,260,284,285,

now, as n=25. Median =(n+1)/2 th observation = (25+1)/2= 13th observation

Therefore, Median=229 seconds

(c) Variance= $[(\sum x_{i}-\bar{x})^{2}/(n-1)]$ ={(1296330)-25*(225.28)^2}/24=(1296300-126877)/24=1148.043 seconds

(d) Standard deviation=$\sqrt{[(\sum x_{i}-\bar{x})^{2}/(n-1)]}$= $\sqrt{Var(x)}$=$\sqrt{1148.043}$=33.88279

For mean in minutes, we can either convert our entire data into minutes by dividing each observation by 60. Our use the generalized formulas given in the image attached.

Here c=60

(e) Mean in minutes=$\sum x_{i}/(n*c)$=(5632/(25*60))=225.280/60=3.755 in minutes

(f) Median in minutes= Median in seconds/c=229/60=3.817 in minutes

(g)Variance in Minutes=$[(\sum x_{i}-\bar{x})^{2}/((n-1)*c^{2})]$={(1296330)-25*(225.28)^2}/(24*60*60)=(1296300-126877)/(24*3600)=1148.043/3600 =0.318901

(h) Standard deviation=$\sqrt{[(\sum x_{i}-\bar{x})^{2}/((n-1)*c^{2})]}$= $\sqrt{Var(x)/c^{2}}$ =$\sqrt{1148.043/3600}$=33.88279/60=0.564713

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