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Mathematics

1 answer:

Softa [21]3 years ago

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Answer:

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The manager of a bank recorded the amount of time each customer spent waiting in line during peak business hours one Monday. The

zavuch27 [327]

The standard deviation of the frequency distribution is 5.54

<h3>How to determine standard deviation?</h3>

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:

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