Respuesta :
Answer:
0.14918.
Step-by-step explanation:
We have been given that a particular fruit's weights are normally distributed, with a mean of 610 grams and a standard deviation of 31 grams. We are asked to find the probability that a fruit will weigh between 589 grams and 602 grams.
First of all, we will find z-score corresponding to 589 and 602 using z-score formula.
[tex]z=\frac{x-\mu}{\sigma}[/tex]
[tex]z=\frac{589-610}{31}=\frac{-21}{31}=-0.677\approx -0.68[/tex]
[tex]z=\frac{602-610}{31}=\frac{-8}{31}=-0.258067\approx -0.26[/tex]
Now, we need to find [tex]P(-0.68<z<-0.26)[/tex].
Using property [tex]P(a<z<b)=P(z<b)-P(z<a)[/tex], we will get:
[tex]P(-0.68<z<-0.26)=P(z<-0.26)-P(z<-0.68)[/tex]
From normal distribution table, we will get:
[tex]P(-0.68<z<-0.26)=0.39743-0.24825[/tex]
[tex]P(-0.68<z<-0.26)=0.14918[/tex]
Therefore, the probability that one fruit will weight between 589 grams and 602 grams is 0.14918.
