Answer :
The percentage of women who have weights that are within those limits are 49.97%.
Yes, many women were excluded with those specifications.
How to determine the required percentage?
In Mathematics and Statistics, the z-score of a given sample size or data set can be calculated by using the following formula:
Z-score, z = (x - μ)/σ
Where:
- σ represents the standard deviation.
- x represents the sample score.
- μ represents the mean score.
For a weight of 147 lb, the z-score is given by:
Z-score, z = (147 - 173.7)/43.2
Z-score, z = -26.7/43.2
Z-score, z = -0.62.
For a weight of 205 lb, the z-score is given by:
Z-score, z = (205 - 173.7)/43.2
Z-score, z = 31.3/43.2
Z-score, z = 0.73.
By using the standardized normal distribution table, the percentage of women who have weights between 147 lb and 205 lb can be calculated as follows:
P(-0.62 ≤ z ≤ 0.73) = P(z ≤ 0.73) - P(z ≤ -0.62)
P(z ≤ 0.73) - P(z ≤ -0.62) = (0.7673 - 0.2676) × 100
P(z ≤ 0.73) - P(z ≤ -0.62) = 0.4997 × 100
Percentage = 49.97%.
For the percentage of women who are excluded from this limit, we have:
The percentage excluded = 100% - 49.97%.
The percentage excluded = 50.03%.
In this context, we can conclude that many women were excluded with those specifications.
Read more on z-score here: brainly.com/question/26714379
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