Question

The lengths of fish of a certain type have a normal distribution with mean 38 cm. It is found that 5% of the fish are longer than 50 cm.

(ii) When fish are chosen for sale, those shorter than 30 cm are rejected. Find the proportion of fish rejected.

Answer 1

Do you have a z-table on hand? Look at the z-score for the upper 5%.

Answer 2

I have a z table, yes.

Answer 3

Alternatively, if you want to try \(\left[\frac{2}{\sqrt\pi}\int_{z\sigma}^{\infty}e^{-t^2}dt\right]_{t=\frac{n}{\sqrt2}}\).

Answer 4

Oh, okay. Well, what's the z-score for the upper 5%?

Answer 5

What do you mean upper?

Answer 6

Do you know what a z-table does?

Answer 7

Yes, I do. So, when it's >0.05, |z| will be 1.645...

Answer 8

But I don't know what you mean by upper?

Answer 9

Oh, nothing. Throwing random terms in. Alright, so \(1.645\sigma\) and above are longer than \(50\text{cm}\)--in other words, \(12\text{cm}\) above average. That means \(1.645\sigma=12\). What is \(\sigma\)?

Answer 10

Also, I need to sleep. Good night; sorry I can't stay. :( Good luck.

Answer 11

\[\sigma= 7.29\]

Answer 12

and after that?

Answer 13

Perhaps this website can help you a bit..
http://www.regentsprep.org/Regents/math/algtrig/ATS2/NormalLesson.htm

Answer 14

find z-score for 30cm

--> z = (30-50)/7.29

Answer 15

oops i meant 38 not 50

Answer 16

Which is -1.0974. Then what?

Answer 17

look it up in table and that will tell you the percentage of fish that are rejected

Answer 18

Thanks! :D I have a second part of the question

Answer 19

9 fish are chosen at random. Find the probability that at least one of them is longer than 50 cm

Answer 20

you know that probability of being longer than 50 is 5% right?

now you have to use binomial distribution:
P(none are >50) = (.95)^9
P(at least one) = 1 - (.95)^9

Answer 21

Thanks! :D