Question

The Acme Company manufactures widgets. The distribution of widget weights is bell-shaped. The widget weights have a mean of 46 ounces and a standard deviation of 3 ounces. Use the Standard Deviation Rule, also known as the Empirical Rule.
Suggestion: sketch the distribution in order to answer these questions.


a. 68% of the widget weights lie between ________ and________

b. What percentage of the widget weights lie between 43 and 87 ounces?

c. What percentage of the widget weights lie below 76 ?

Answer 1

Answer:

(a) 68% of the widget weights lie between 43 ounces and 49 ounces.

(b) The percentage of the widget weights lie between 43 and 87 ounces is 15.87%.

(c) The percentage of the widget weights lie below 76 is 100%.

Step-by-step explanation:

Let X = weight of widgets manufactured by Acme Company.

The distribution of the random variable X is, N (μ = 46, σ² = 3²).

According to the Empirical Rule in a normal distribution with mean µ and standard deviation σ, nearly all the data will fall within 3 standard deviations of the mean. The empirical rule can be broken into three parts:

• 68% data falls within 1 standard deviation of the mean.                       That is P (µ - σ ≤ X ≤ µ + σ) = 0.68.
• 95% data falls within 2 standard deviations of the mean.                   That is P (µ - 2σ ≤ X ≤ µ + 2σ) = 0.95.
• 99.7% data falls within 3 standard deviations of the mean.                   That is P (µ - 3σ ≤ X ≤ µ + 3σ) = 0.997.

(a)

According to the Empirical rule, 68% data falls within 1 standard deviation of the mean.

P (µ - σ ≤ X ≤ µ + σ) = 0.68.

Compute the upper and lower values as follows:

µ - σ = 46 - 3 = 43 ounces

µ + σ = 46 + 3 = 49 ounces

Thus, 68% of the widget weights lie between 43 ounces and 49 ounces.

(b)

Compute the probability of the widget weights lie between 43 and 87 ounces as follows:

$P(43$

                          $=P(-1$

*Use a z-table.

The percentage is, 0.1587 × 100 = 15.87%.

Thus, the percentage of the widget weights lie between 43 and 87 ounces is 15.87%.

(c)

Compute the probability of the widget weights lie below 76 as follows:

$P(X$

                  $=P(Z$

*Use a z-table.

The percentage is, 1 × 100 = 100%.

Thus, the percentage of the widget weights lie below 76 is 100%.