Question

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?

Answer 1

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 %

Step-by-step 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

$Z = \frac{x-mean}{S.D} = \frac{43-50}{10} = -0.7$

Given X = 87

$Z = \frac{x-mean}{S.D} = \frac{87-50}{10} = 3.7$

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 %