Thank you for visiting A survey found that women s heights are normally distributed with a mean of 62 8 inches and a standard deviation of 3 7 inches. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!
Answer :
The percentage of men meeting the height requirement is 12.10%.
To find the percentage of men meeting the height requirement, we need to calculate the probability that a randomly selected man's height falls within the range of 57 inches to 64 inches.
First, we need to standardize the height range using the Z-score formula:
[tex]\[ Z = \frac{X - \mu}{\sigma} \][/tex]
where:
- X is the height requirement (either 57 inches or 64 inches),
- [tex]\( \mu \)[/tex] is the mean height of men (68.1 inches), and
- [tex]\( \sigma \)[/tex] is the standard deviation of men's heights (3.5 inches).
For the lower limit (57 inches):
[tex]\[ Z_{\text{lower}} = \frac{57 - 68.1}{3.5} = \frac{-11.1}{3.5} \approx -3.17 \][/tex]
For the upper limit (64 inches):
[tex]\[ Z_{\text{upper}} = \frac{64 - 68.1}{3.5} = \frac{-4.1}{3.5} \approx -1.17 \][/tex]
Next, we need to find the probabilities corresponding to these Z-scores using a standard normal distribution table (Z-table).
The probability of a Z-score being less than -3.17 is very close to 0, and the probability of a Z-score being less than -1.17 is approximately 0.1210.
To find the percentage of men meeting the height requirement, we need to find the probability of the range between these Z-scores (from -3.17 to -1.17) using the cumulative probability from the Z-table.
We can do this by finding the probability of a Z-score less than -1.17 and subtracting the probability of a Z-score less than -3.17 from it:
[tex]\[ P(-3.17 < Z < -1.17) = P(Z < -1.17) - P(Z < -3.17) \][/tex]
Now, we find these probabilities from the Z-table:
- [tex]\( P(Z < -1.17) \approx 0.1210 \)[/tex] (from earlier calculation)
- [tex]\( P(Z < -3.17) \approx 0 \)[/tex] (very close to 0)
Therefore, the percentage of men meeting the height requirement is approximately [tex]\(0.1210 \times 100\% = 12.10\% \)[/tex].
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