A survey found that women's heights are normally distributed with mean 63.9 in and standard deviation 2.4in. A branch of the military requires women's heights to be between 58in and 80 in . Fund percentage of women who meets the height requirements

1 answer

To find the percentage of women who meet the height requirements of being between 58 inches and 80 inches, we can use the properties of the normal distribution.

Given:

  • Mean (μ) = 63.9 inches
  • Standard deviation (σ) = 2.4 inches

We will calculate the z-scores for both height requirements:

  1. For 58 inches: \[ z = \frac{(X - \mu)}{\sigma} \] \[ z_{58} = \frac{(58 - 63.9)}{2.4} = \frac{-5.9}{2.4} \approx -2.46 \]

  2. For 80 inches: \[ z_{80} = \frac{(80 - 63.9)}{2.4} = \frac{16.1}{2.4} \approx 6.71 \]

Next, we can look up the z-scores in the standard normal distribution table or use a calculator or statistical software to find the corresponding probabilities.

  • For \( z = -2.46 \):

    • The probability is approximately \( P(Z < -2.46) \approx 0.0069 \) (or 0.69%).
  • For \( z = 6.71 \):

    • The probability is approximately \( P(Z < 6.71) \approx 1.0000 \) (or 100%). This is a very high z-score, indicating virtually all women fall below this height.

Now we find the percentage of women whose heights lie between 58 inches and 80 inches. This is computed as follows: \[ P(58 < X < 80) = P(X < 80) - P(X < 58) \] Since \( P(X < 80) \approx 1.0000 \) and \( P(X < 58) \approx 0.0069 \): \[ P(58 < X < 80) \approx 1.0000 - 0.0069 \approx 0.9931 \]

Thus, the percentage of women who meet the height requirements is: \[ 0.9931 \times 100 \approx 99.31% \]

Therefore, approximately 99.31% of women meet the military height requirements.