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?

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 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 82 pounds and a standard deviation of 9.8 pounds.

This means that

What percent of the panthers weigh less than 76 pounds?

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

has a pvalue of 0.2709

0.2709*100% = 27.09%

27.09% of the panthers weigh less than 76 pounds

i'm pretty sure it's 477,585,296, but perhaps my math is incorrect.

were is the equation