Answer:
(a) 68% of the widget weights lie between <u>43 ounces</u> and <u>49 ounces</u>.
(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 <em>X</em> = weight of widgets manufactured by Acme Company.
The distribution of the random variable <em>X</em> is, N (<em>μ </em>= 46, <em>σ</em>²<em> </em>=<em> </em>3²).
According to the Empirical Rule in a normal distribution with mean <em>µ</em> and standard deviation <em>σ</em>, 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:
<em>µ</em> - <em>σ</em> = 46 - 3 = 43 ounces
<em>µ</em> + <em>σ</em> = 46 + 3 = 49 ounces
Thus, 68% of the widget weights lie between <u>43 ounces</u> and <u>49 ounces</u>.
(b)
Compute the probability of the widget weights lie between 43 and 87 ounces as follows:
*Use a <em>z</em>-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:
*Use a <em>z</em>-table.
The percentage is, 1 × 100 = 100%.
Thus, the percentage of the widget weights lie below 76 is 100%.
