Question

Apple weights are normally distriuted with a mean 0.33 pounds and and standard deviation of 0.06 pounds. Jermaine went to the store and found the an apple that weighed 0.25 pounds. What percent of the apples weigh less than Jermaine's apple? % (write your answer as a percent WITHOUT the percent sign and with two numbers after the decimal point)

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Answer 1

Using the normal distribution, it is found that 10.56% of the apples weigh less than Jermaine's apple.

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In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

• It measures how many standard deviations the measure X 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, which is the percentile of X, that is, the proportion of measures that are less than X.

In this problem:

• Mean of 0.33 pounds, thus $\mu = 0.33$.
• Standard deviation of 0.06 pounds, thus $\sigma = 0.06$.
• The proportion that is less than 0.25 pounds is the p-value of Z when X = 0.25, thus:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{0.25 - 0.33}{0.06}$

$Z = -1.25$

$Z = -1.25$ has a p-value of 0.1056.

0.1056 x 100% = 10.56%.

10.56% of the apples weigh less than Jermaine's apple.

A similar problem is given at https://brainly.com/question/13411796