The heights of women follow a normal distribution with a mean of 64.5 inches and a standard deviation of 2.8 inches. (a) (2 points) What percent of women are taller than 72 inches (6 feet)?
(b) (4 points) From a SRS of 50 women, what is the probability that their mean height is between 60 inches and 72 inches?
The Correct Answer and Explanation is :
To solve these questions, we’ll use properties of the normal distribution.
(a) What percent of women are taller than 72 inches (6 feet)?
We are given:
- Mean height = 64.5 inches
- Standard deviation = 2.8 inches
We need to find the probability that a randomly selected woman is taller than 72 inches. This can be done by calculating the Z-score for 72 inches and then finding the corresponding probability.
The Z-score formula is:
[
Z = \frac{X – \mu}{\sigma}
]
Where:
- ( X ) is the value (72 inches),
- ( \mu ) is the mean (64.5 inches),
- ( \sigma ) is the standard deviation (2.8 inches).
Substituting the given values:
[
Z = \frac{72 – 64.5}{2.8} = \frac{7.5}{2.8} \approx 2.68
]
Now, we need to find the probability that a woman is taller than 72 inches. Using a standard normal distribution table or a Z-table, we look up the cumulative probability for a Z-score of 2.68. The cumulative probability for a Z-score of 2.68 is approximately 0.9963. This represents the probability of being less than 72 inches.
To find the probability of being greater than 72 inches, we subtract this value from 1:
[
P(X > 72) = 1 – 0.9963 = 0.0037
]
So, approximately 0.37% of women are taller than 72 inches.
(b) From a SRS (Simple Random Sample) of 50 women, what is the probability that their mean height is between 60 inches and 72 inches?
When we are dealing with a sample mean, we need to use the standard error of the mean (SEM). The formula for SEM is:
[
\text{SEM} = \frac{\sigma}{\sqrt{n}}
]
Where:
- ( \sigma ) is the population standard deviation (2.8 inches),
- ( n ) is the sample size (50 women).
Substituting the values:
[
\text{SEM} = \frac{2.8}{\sqrt{50}} \approx \frac{2.8}{7.071} \approx 0.396
]
Now, we want to find the probability that the sample mean height is between 60 and 72 inches. To do this, we first calculate the Z-scores for 60 and 72 inches using the formula:
[
Z = \frac{X – \mu}{\text{SEM}}
]
For ( X = 60 ) inches:
[
Z_1 = \frac{60 – 64.5}{0.396} = \frac{-4.5}{0.396} \approx -11.36
]
For ( X = 72 ) inches:
[
Z_2 = \frac{72 – 64.5}{0.396} = \frac{7.5}{0.396} \approx 18.94
]
Both Z-scores are extremely extreme. The probability corresponding to a Z-score of -11.36 is essentially 0, and the probability for a Z-score of 18.94 is essentially 1. Therefore, the probability that the sample mean height is between 60 and 72 inches is approximately 1.
Summary:
- (a) Approximately 0.37% of women are taller than 72 inches.
- (b) The probability that the mean height of a sample of 50 women is between 60 and 72 inches is essentially 100% (or probability 1).