Question

To study the output of a machine that fills boxes with cereal, a quality control engineer weighed 150 boxes of Brand X cereal. The frequency distribution in the following table summarizes his findings. Find the standard deviation of the weight of the boxes of cereal. (Round your answer to three decimal places.)
Incorrect: Your answer is incorrect.
oz
x = Weight
(ounces) Number of
boxes
15.3 ≤ x < 15.6 14
15.6 ≤ x < 15.9 25
15.9 ≤ x < 16.2 84
16.2 ≤ x < 16.5 18
16.5 ≤ x < 16.8 9

Answer 1

Answer:

$\sigma = 0.07770$

Step-by-step explanation:

Given

The above table

Required

The standard deviation

First, calculate the class mid-point. This is the average of the class interval.

The mid-point x is:

$x_1 = \frac{1}{2}(15.3 + 15.6) =\frac{1}{2} * 30.9 = 15.45$

$x_2 = \frac{1}{2}(15.6 + 15.9) =\frac{1}{2} * 31.5 = 15.75$

$x_3 = \frac{1}{2}(15.9 + 16.2) =\frac{1}{2} * 32.1 = 16.05$

$x_4 = \frac{1}{2}(16.2 + 16.5) =\frac{1}{2} * 32.7 = 16.35$

$x_5 = \frac{1}{2}(16.5 + 16.8) =\frac{1}{2} * 33.3 = 16.65$

So, we have:

$\begin{array}{cccccc}x & {15.45} & {15.75} & {16.05} & {16.35} & {16.65} \ \\ f & {14} & {25} & {84} & {18} & {9} \ \end{array}$

Calculate mean

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

$\bar x = \frac{15.45 * 14 + 15.75 * 25 + 16.05 * 84 + 16.35 * 18 + 16.65 * 9}{14 + 25 + 84 + 18 + 9}$

$\bar x = \frac{2402.4}{150}$

$\bar x = 16.016$

The standard deviation is:

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

$\sigma = \sqrt{\frac{(15.45 - 16.016)^2+ (15.75 - 16.016)^2+ (16.05 - 16.016)^2+ (16.35 - 16.016)^2+ (16.65 - 16.016)^2}{14 + 25 + 84 + 18 + 9}}$

$\sigma = \sqrt{\frac{0.90578}{150}}$

$\sigma = \sqrt{0.00603853333}$

$\sigma = 0.07770$