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:

[tex]\sigma = 0.07770[/tex]

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:

[tex]x_1 = \frac{1}{2}(15.3 + 15.6) =\frac{1}{2} * 30.9 = 15.45[/tex]

[tex]x_2 = \frac{1}{2}(15.6 + 15.9) =\frac{1}{2} * 31.5 = 15.75[/tex]

[tex]x_3 = \frac{1}{2}(15.9 + 16.2) =\frac{1}{2} * 32.1 = 16.05[/tex]

[tex]x_4 = \frac{1}{2}(16.2 + 16.5) =\frac{1}{2} * 32.7 = 16.35[/tex]

[tex]x_5 = \frac{1}{2}(16.5 + 16.8) =\frac{1}{2} * 33.3 = 16.65[/tex]

So, we have:

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

Calculate mean

[tex]\bar x = \frac{\sum fx}{\sum f}[/tex]

[tex]\bar x = \frac{15.45 * 14 + 15.75 * 25 + 16.05 * 84 + 16.35 * 18 + 16.65 * 9}{14 + 25 + 84 + 18 + 9}[/tex]

[tex]\bar x = \frac{2402.4}{150}[/tex]

[tex]\bar x = 16.016[/tex]

The standard deviation is:

[tex]\sigma = \sqrt{\frac{\sum(x_i - \bar x)^2}{\sum f}}[/tex]

[tex]\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}}[/tex]

[tex]\sigma = \sqrt{\frac{0.90578}{150}}[/tex]

[tex]\sigma = \sqrt{0.00603853333}[/tex]

[tex]\sigma = 0.07770[/tex]

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