Answer: 0.0660
Step-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.
