The manager of a bank recorded the amount of time each customer spent waiting in

Question

The manager of a bank recorded the amount of time each customer spent waiting in line during peak business hours one Monday. The frequency distribution below summarizes the results. Find the sample mean, sample variance, and sample standard deviationof the waiting time. Round your answer to one decimal place. (Mean =5 points)

(Variance 5 points) (Standard deviation = 3 points)

Waiting time   (minutes)

Number of customers

0 - 3

15

4 - 7

9

8 - 11

11

12 - 15

6

16 - 19

5

20 - 23

3

24 - 27

1

Waiting time   (minutes)

Number of customers

0 - 3

15

4 - 7

9

8 - 11

11

12 - 15

6

16 - 19

5

20 - 23

3

24 - 27

1

Explanation / Answer

Answer:

Mean = 7.2

Standard Deviation = 6.664

Variance = 44.408

Occurance(X) Frequency(f) Freq*X (X-mean) (X-mean)2 f*(X-mean)2 0 - 3 15 0 -7.2 51.84 777.6 4 - 7 9 36 -3.2 10.24 92.16 8 - 11 11 88 0.8 0.64 7.04 12 - 15 6 72 4.8 23.04 138.24 16 - 19 5 80 8.8 77.44 387.2 20 - 23 3 60 12.8 163.84 491.52 24 - 27 1 24 16.8 282.24 282.24 Total -> 50 360 - - 2176

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