## 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”]

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3 months 2022-02-10T14:06:43+00:00 1 Answer 0 views 0

1. ## Answer with explanation:

Given : The distribution of the lengths of a commercially caught fish is bell-shapped  ( i.e. normally distributed ) with $$\mu=24\ cm$$ and $$\sigma=4\ cm$$.

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 :

$$P(x<20)=P(\dfrac{x-\mu}{\sigma}<\dfrac{20-24}{4})\\\\=P(z<-1)\ \ [\because z=\dfrac{x-\mu}{\sigma}]\\\\=1-P(z<1)\ \ [\because\ P(Z<-z)=1-P(Z<z)]\\\\=1-0.8413\ \ [\text{By z-table}]\\\\ =0.1587\approx0.16$$

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 :

$$P(20<x<32)=P(\dfrac{20-24}{4}<\dfrac{x-\mu}{\sigma}<\dfrac{32-24}{4})\\\\=P(-1<x<2)\\\\=P(x<2)-P(x<-1)\\\\=0.9772- 0.1587\ [\text{By (1) and using z-table}]=0.8185\approx\ 0.82$$

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”.