Respuesta :
Answer with explanation:
Given : The distribution of the lengths of a commercially caught fish is bell-shapped ( i.e. normally distributed ) with [tex]\mu=24\ cm[/tex] and [tex]\sigma=4\ cm[/tex].
Let x be the length of a commercially caught fish .
Any fish measuring less than 20 cm must be released.
1) The probability that a fish measuring less than 20 cm :
[tex]P(x<20)=P(\dfrac{x-\mu}{\sigma}<\dfrac{20-24}{4})\\\\=P(z<-1)\ \ [\because z=\dfrac{x-\mu}{\sigma}]\\\\=1-P(z<1)\ \ [\because\ P(Z<-z)=1-P(Z<z)]\\\\=1-0.8413\ \ [\text{By z-table}]\\\\ =0.1587\approx0.16[/tex]
∴ The proportion of fish that must be released is about 0.16.
2) The probability that a fish measuring between 20 cm and 32 cm :
[tex]P(20<x<32)=P(\dfrac{20-24}{4}<\dfrac{x-\mu}{\sigma}<\dfrac{32-24}{4})\\\\=P(-1<x<2)\\\\=P(x<2)-P(x<-1)\\\\=0.9772- 0.1587\ [\text{By (1) and using z-table}]=0.8185\approx\ 0.82[/tex]
If total 500 fish are caught, then number of fish have lengths between 20 cm and 32 cm = 0.8185 x (500)=409.25 ≈409
The nearest option from the given options is "408".