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 deviation of the waiting time. Round your answer to one decimal place.Waiting time0-3,4-7,8-11,12,15,16-19,20-23,24-27Number of customers15,9,11,6,5,3,1

Answer 1

Answer:

4.517

Step-by-step explanation:

Given the data:

______m __ f___fx __f(m - x)²

0-4 __ 2___11__22 __ 385.51

4-8 __ 6___15__90__ 55.29

8-12 _ 10__ 15__150__64.896

12-16_ 14___8__112__295.73

16-20_18__ 0__ 0 ___0

20-24_22__1__ 22 ___198.25

Mean (x) = Σfx / Σf

Mean = (22 + 90 + 150 + 112 + 0 + 22) / (11 + 15 + 15 + 8 + 0 + 1)

= 396 / 50

= 7.92

Answer 2

The standard deviation will be "6.597".

According to the question,

C.I                   f                     $x_i$                      $f_i x_i$             $f_i(x_i- \bar{x})^2$

0-3                 15                  1.5                    22.5               777.6

4-7                  9                   5.5                   49.5               92.16    

8-11                 11                   9.5                   104.5              7.04

12-15               6                   13.5                    81                138.24  

16-19               5                   17.5                   87.5              387.2

20-23             3                   21.5                   64.5              491.52    

24-27              1                    25.5                  25.5             282.24    

                                                                       435                2176  

Mean,

= $\frac{\Sigma f_i x_i}{\Sigma f_1}$

= $\frac{453}{50}$

= $8.7$

Variance,

= $\frac{1}{N} \Sigma f_i(x_i- \bar x)^2$

= $\frac{2176}{50}$

= $43.52$

and,

Standard deviation,

= $\sqrt{variance}$

= $\sqrt{43.52}$

= $6.597$

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https://brainly.com/question/18801702