Question

1. men’s heights the distribution of heights of adult american men is approximately normal with mean 69 inches and standard deviation 2.5 inches. draw an accurate sketch of the distribution of men’s heights. be sure to label the mean, as well as the points 1, 2, and 3 standard deviations away from the mean on the horizontal axis.
2. potato chips the distribution of weights of 9 - ounce bags of a particular brand of potato chips is approximately normal with mean μ = 9.12 ounces and standard deviation σ = 0.05 ounce. draw an accurate sketch of the distribution of potato chip bag weights. be sure to label the mean, as well as the points 1, 2, and 3 standard deviations away from the mean on the horizontal axis.
3. men’s heights refer to exercise 41. use the 68 - 95 - 99.7 rule to answer the following questions. show your work!

(a) between what heights do the middle 95% of men fall?
(b) what percent of men are taller than 74 inches?
(c) what percent of men are between 64 and 66.5 inches tall?
(d) a height of 71.5 inches corresponds to what percentile of adult male american heights?

Explanation:

Step1: Recall the 68 - 95 - 99.7 rule

The 68 - 95 - 99.7 rule for a normal distribution states that about 68% of the data lies within 1 standard deviation of the mean ($\mu\pm\sigma$), about 95% lies within 2 standard deviations of the mean ($\mu\pm2\sigma$), and about 99.7% lies within 3 standard deviations of the mean ($\mu\pm3\sigma$). The mean of men's heights is $\mu = 69$ inches and the standard deviation is $\sigma=2.5$ inches.

Step2: Answer part (a)

The middle 95% of the data in a normal distribution lies within 2 standard deviations of the mean.
Lower bound: $\mu - 2\sigma=69-2\times2.5=69 - 5=64$ inches.
Upper bound: $\mu + 2\sigma=69 + 2\times2.5=69+5 = 74$ inches.

Step3: Answer part (b)

First, find the z - score. The z - score formula is $z=\frac{x-\mu}{\sigma}$. For $x = 74$ inches, $z=\frac{74 - 69}{2.5}=\frac{5}{2.5}=2$.
The total area under the normal curve is 1. The area to the left of $z = 2$ is about 0.9772 (from the standard normal table or the 68 - 95 - 99.7 rule, since 95% of the data is within $z=- 2$ and $z = 2$, so the area to the left of $z = 2$ is $0.5+\frac{0.95}{2}=0.9772$).
The area to the right of $z = 2$ (the percentage of men taller than 74 inches) is $1 - 0.9772=0.0228$ or 2.28%.

Step4: Answer part (c)

For $x_1 = 64$ inches, $z_1=\frac{64 - 69}{2.5}=\frac{-5}{2.5}=-2$.
For $x_2 = 66.5$ inches, $z_2=\frac{66.5 - 69}{2.5}=\frac{-2.5}{2.5}=-1$.
The area to the left of $z=-1$ is about 0.1587 and the area to the left of $z=-2$ is about 0.0228.
The area between $z=-2$ and $z=-1$ is $0.1587 - 0.0228=0.1359$ or 13.59%.

Step5: Answer part (d)

For $x = 71.5$ inches, $z=\frac{71.5 - 69}{2.5}=\frac{2.5}{2.5}=1$.
The area to the left of $z = 1$ is about 0.8413 (since 68% of the data is within $z=-1$ and $z = 1$, so the area to the left of $z = 1$ is $0.5+\frac{0.68}{2}=0.8413$). So a height of 71.5 inches corresponds to the 84.13th percentile.

Answer:

(a) 64 inches and 74 inches
(b) 2.28%
(c) 13.59%
(d) 84.13th percentile