A biologist found the wingspans of a group of monary butterflies to be normally distributed with a mean of 52.5 mm and a standard deviation of 2.5 mm. What percent of the butterflies had the following wingspans? A. Less than 48.9 mm B. Between 49 and 55

A. 7.5%

B. 68.26%

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean and standard deviation , the z-score of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Mean of 52.5 mm and a standard deviation of 2.5 mm.

This means that

A. Less than 48.9 mm

The proportion is the pvalue of Z when X = 48.9. So

has a pvalue of 0.075

0.075*100% = 7.5%, which is the answer.

B. Between 49 and 55

The proportion is the pvalue of Z when X = 55 subtracted by the pvalue of Z when X = 49. So

X = 55

has a pvalue of 0.8413

X = 49

has a pvalue of 0.1587

0.8413 - 0.1587 = 0.6826

0.6826*100% = 68.26%, which is the answer.

the answer is d. 75%

1) is right

step-by-step explanation: