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The Acme Company manufactures widgets. The distribution of widget weights is bell-shaped. The widget weights have a mean of 50 ounces and a standard deviation of 10 ounces. Use the Empirical Rule, what percentage of the widget weights lie between 43 and 87 ounces?

User Larina
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1 Answer

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Answer:

The probability of the widget weights lie between 43 and 87 ounces

P(43 ≤ x ≤ 87 ) = 0.7605

The percentage of the widget weights lie between 43 and 87 ounces

P(43 ≤ x ≤ 87 ) = 76 %

Explanation:

Step(i):-

Given mean of the Population = 50ounces

Given standard deviation of the Population = 10 ounces

Let 'X' be the random variable in Normal distribution

Given X = 43


Z = (x-mean)/(S.D) = (43-50)/(10) = -0.7

Given X = 87


Z = (x-mean)/(S.D) = (87-50)/(10) = 3.7

Step(ii):-

The probability of the widget weights lie between 43 and 87 ounces


P(43 \leq x\leq 87 ) = P(-0.7\leq z\leq 3.7 )

= |A(3.7) + A(-0.7)|

= |A(3.7) + A(0.7)| (∵A(-z) = A(z))

= 0.4994 + 0.2611

= 0.7605

Final answer:-

The probability of the widget weights lie between 43 and 87 ounces

P(43 ≤ x ≤ 87 ) = 0.7605

The percentage of the widget weights lie between 43 and 87 ounces

P(43 ≤ x ≤ 87 ) = 76 %

User ZooMMX
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