Question

A particular fruit's weights are normally distributed, with a mean of 353 grams and a standard deviation of 6 grams. If you pick one fruit at random, what is the probability that it will weigh between 334 grams and 344 grams?

Answer 1

Answer: 0.0660

Explanation:

Given : A particular fruit's weights are normally distributed with

Mean :
`\mu=353\text{ grams}`

Standard deviation :
`\sigma=6\text{ grams}`

The formula to calculate the z-score is given by :-

`z=(x-\mu)/(\sigma)`

Let x be the weight of randomly selected fruit.

Then for x = 334 , we have

`z=(334-353)/(6)=-3.17`

for x = 344 , we have

`z=(344-353)/(6)=-1.5`

The p-value :
`P(334<x<353)=P(-3.17<z<-1.5)`

`P(-1.5)-P(-3.17)=0.0668072-0.000771=0.0660362\approx0.0660`

Thus, the probability that it will weigh between 334 grams and 344 grams = 0.0660.