## The distribution of the lengths of a commercially caught fish is bell-shapped with mean 24 cm and standard deviation 4 cm.1- Any fish measur

Question

The distribution of the lengths of a commercially caught fish is bell-shapped with mean 24 cm and standard deviation 4 cm.1- Any fish measuring less than 20 cm must be released. the proportion of fish that must be released is about [ Select ] [“.05”, “.025”, “.16”, “.68”, “.32”] 2- 500 fish are caught. Approximately how many of them have lengths between 20 cm and 32 cm? [ Select ] [“340”, “25”, “408”, “80”, “480”]

in progress
0

Math
3 months
2022-02-10T14:06:43+00:00
2022-02-10T14:06:43+00:00 1 Answer
0 views
0
## Answers ( )

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”.