Respuesta :
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:
[tex]P(43<X<87) =P(\frac{43-46}{3}<\frac{X-\mu}{\sigma}<\frac{87-46}{3})[/tex]
[tex]=P(-1<Z<41)\\=P(Z<41)-P(Z<-1)\\=1-0.8413\\=0.1587[/tex]
*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:
[tex]P(X<76)=P(\frac{X-\mu}{\sigma}<\frac{76-46}{3})[/tex]
[tex]=P(Z<10)\\\approx1[/tex]
*Use a z-table.
The percentage is, 1 × 100 = 100%.
Thus, the percentage of the widget weights lie below 76 is 100%.
