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?
Accepted Solution
A:
Answer: 0.0660Step-by-step explanation:Given : A particular fruit's weights are normally distributed with Mean : [tex]\mu=353\text{ grams}[/tex]Standard deviation : [tex]\sigma=6\text{ grams}[/tex]The formula to calculate the z-score is given by :-[tex]z=\dfrac{x-\mu}{\sigma}[/tex]Let x be the weight of randomly selected fruit.Then for x = 334 , we have [tex]z=\dfrac{334-353}{6}=-3.17[/tex]for x = 344 , we have [tex]z=\dfrac{344-353}{6}=-1.5[/tex]The p-value : [tex]P(334<x<353)=P(-3.17<z<-1.5)[/tex][tex]P(-1.5)-P(-3.17)=0.0668072-0.000771=0.0660362\approx0.0660[/tex]Thus, the probability that it will weigh between 334 grams and 344 grams = 0.0660.