Question

To study the efficiency of its new price-scanning equipment, a local supermarket monitored the amount of time its customers had to wait in line. The frequency distribution in the following table summarizes the findings. Find the standard deviation of the amount of time spent in line. (Round your answer to three decimal places.)
______Min


x = Time (minutes) Number of Customers
0 ≤ x < 1 75
1 ≤ x < 2 58
2 ≤ x < 3 64
3 ≤ x < 4 40
4 ≤ x < 5 38

I tried
Frequency Midpoint (xm) f*xm (f*xm)^2
75 .5 37.5 140625
58 1.5 87 7569
64 2.5 160 25600
40 3.5 140 19600
38 4.5 171 29241
+------ +------ +-------
275 595.5 83416.25

(f*xm)^2= (595.5)^2 = 354620.25
s^2= (1/n-1)[EX^2-(((EX)^2)/n)]
(1/274)[83416.25-((565.5^2)/275)]
(1/274)[83416.25-((354620.25)/275)]
(1/274)[83416.25-(1289.528182)]
(1/274)[82126.72182]= 299.7325614
Variance =299.7325614
Standard Deviation=Square Root of 299.7325614
Which equals 17.31278607
Rounded three decimal places= 17.313
However this answer was incorrect.
Please help!

Answer 1

This is to help the spacing
x = Time (minutes) Number of Customers
0 ≤ x < 1………………………………….. 75
1 ≤ x < 2………………………………….. 58
2 ≤ x < 3…………………………………..64
3 ≤ x < 4………………………………….. 40
4 ≤ x < 5………………………………….. 38

I tried
Frequency Midpoint (xm) f*xm (f*xm)^2
75—————-.5———— 37.5——140625
58—————-1.5———— 87 —— 7569
64—————-2.5———— 160 ——25600
40—————-3.5———— 140——19600
38—————-4.5 ———— 171 ——29241
+------ +------ +-------
275——————————595.5-—83416.25