The weight of adult female panthers in one habitat area are normally distributed with a mean of 82 pounds and a standard deviation of 9.8 pounds. What percent of the panthers weigh less than 76 pounds?

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Answer:

27.09% of the panthers weigh less than 76 pounds

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 82 pounds and a standard deviation of 9.8 pounds.

This means that [tex]\mu = 82, \sigma = 9.8[/tex]

What percent of the panthers weigh less than 76 pounds?

The proportion is the pvalue of Z when X = 76. So

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

[tex]Z = \frac{76 - 82}{9.8}[/tex]

[tex]Z = -0.61[/tex]

[tex]Z = -0.61[/tex] has a pvalue of 0.2709

0.2709*100% = 27.09%

27.09% of the panthers weigh less than 76 pounds

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