Question

A particular fruit's weights are normally distributed, with a mean of 275 grams and a standard deviation of 19 grams. If you pick one fruit at random, what is the probability that it will weigh between 244 grams and 305 grams?

Answer 1

Answer:

0.8913 = 89.13% probability that it will weigh between 244 grams and 305 grams.

Step-by-step explanation:

Normal Probability Distribution:

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

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

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

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 275 grams and a standard deviation of 19 grams

This means that $\mu = 275, \sigma = 19$

What is the probability that it will weigh between 244 grams and 305 grams?

This is the p-value of Z when X = 305 subtracted by the p-value of Z when X = 244.

X = 305

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

$Z = \frac{305 - 275}{19}$

$Z = 1.58$

$Z = 1.58$ has a p-value of 0.9429.

X = 244

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

$Z = \frac{244 - 275}{19}$

$Z = -1.63$

$Z = -1.63$ has a p-value of 0.0516

0.9429 - 0.0516 = 0.8913

0.8913 = 89.13% probability that it will weigh between 244 grams and 305 grams.