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 standard deviation. Round your answer to one decimal place. Waiting Time (Minutes) Number of Customer 0-3 13 4-7 13 8-11 10 12-15 11 16-19 0 20-23 3

Answer 1

The standard deviation of the frequency distribution is 5.54

How to determine standard deviation?

The table of values is given as:

x     f(x)

0-3 13

4-7  13

8-11 10

12-15 11

16-19 0

20-23 3

Rewrite the table by calculating the class midpoints:

x     f(x)

1.5 13

5.5  13

9.5 10

13.5 11

17.5 0

21.5 3

Start by calculating the mean using:

$\bar x = \frac{\sum fx}{\sum f}$

This gives

$\bar x = \frac{1.5 * 13 + 5.5 * 13 + 9.5 * 10 + 13.5 * 11 + 17.5 * 0 + 21.5 * 3}{13 + 13 + 10 + 11 + 0 +3}$

Evaluate

$\bar x = 7.98$

The standard deviation is then calculated as:

$\sigma = \sqrt{\frac{\sum f(x - \bar x)^2}{\sum f }}$

So, we have:

$\sigma= \sqrt{\frac{13 * (1.5 - 7.98)^2 + 13 * (5.5 - 7.98)^2 + (9.5 - 7.98)^2 * 10 + (13.5 - 7.98)^2 * 11 + (17.5 - 7.98)^2 * 0 + (21.5 - 7.98)^2 * 3}{13 + 13 + 10 + 11 + 0 +3}}$

Evaluate

$\sigma= \sqrt{30.6496}$

Solve

$\sigma= 5.54$

Hence, the standard deviation is 5.54

Read more about standard deviation at:

brainly.com/question/15858152

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