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

Answer:

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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

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 [tex]\mu = 52.5, \sigma = 2.5[/tex]

A. Less than 48.9 mm

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

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{48.9 - 52.5}{2.5}[/tex]

[tex]Z = -1.44[/tex]

[tex]Z = -1.44[/tex] 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

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{55 - 52.5}{2.5}[/tex]

[tex]Z = 1[/tex]

[tex]Z = 1[/tex] has a pvalue of 0.8413

X = 49

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{49 - 52.5}{2.5}[/tex]

[tex]Z = -1[/tex]

[tex]Z = -1[/tex] has a pvalue of 0.1587

0.8413 - 0.1587 = 0.6826

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

The percent of the butterflies had the given wingspans are;

A) 7.5%

B) 76.64%

We are given;

Population mean; μ = 52.5 mm

Population standard deviation; σ = 2.5 mm

Formula for z-score is;

z = (x' - μ)/σ

A) For less than 48.9 mm,P(X' < 48.9);

z = (48.9 - 52.5)/2.5

z = -1.44

From online p-value from z-score calculator, we have;

P(X' < 48.9) = 0.075 or 7.5%

B) For x' between 49 and 55, P(49 < X < 55)

At x' = 49;

z = (49 - 52.5)/2.5

z = -1.4

Z-score at x' = 55;

z = (55 - 52.5)/2.5

z = 1

From online p-value from z-score calculator, the p-value between the two z-scores is;

p = 0.7664 or 76.64%

Read more about z-scores at; https://brainly.com/question/6316394

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