Answer :
Approximately 4.91% of women have weights between 141 and 201 pounds, indicating that very few women are excluded based on those weight specifications.
How many women are within weight limits?
To find the percentage of women with weights within the specified limits, we can calculate the z-scores corresponding to the lower and upper weight limits using the given mean and standard deviation:
Lower z-score = (141 - 167) / 457 = -0.057
Upper z-score = (201 - 167) / 457 = 0.074
Using a standard normal distribution table or a statistical calculator, we can find the probabilities associated with these z-scores:
Lower probability = P(Z < -0.057) = 0.4788
Upper probability = P(Z < 0.074) = 0.5279
To find the percentage of women within the specified weight limits, we subtract the lower probability from the upper probability:
Percentage of women within limits = (0.5279 - 0.4788) * 100 = 4.91%
This means that approximately 4.91% of women have weights between 141 and 201 pounds.
Regarding the question of how many women are excluded with those specifications, we can infer from the low percentage (4.91%) that very few women are excluded based on these weight limits. Therefore, the statement "No, the percentage of women who are excluded, which is equal to the probability found previously, shows that very few women are excluded" is the correct answer (choice A).
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