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.

(i) Find the standard deviation.

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.

(i) Find the standard deviation.

anonymous · Answer

$$z= {d - \mu \over \sigma} $$

callisto · Answer

Oh...
mean + 2sd = 95% of the data
38 + 2sd = 50
2sd = 12
sd = 6
Not sure

anonymous · Answer

$$p(|z|>{50-38 \over \sigma})=0.05???$$

anonymous · Answer

$${50-38 \over \sigma}=0.645$$??

anonymous · Answer

$${12 \over \sigma }= 0.645$$??

anonymous · Answer

$$\sigma = 18.6?$$

callisto · Answer

For a normal distribution curve, the 95% of the data is = mean + 2 s.d.
That means 50 = 38 +2s.d.

anonymous · Answer

I thought that was the formula I had to use though? I mean the one I wrote?

anonymous · Answer

|z|>0.05=
|z|<0.95= 0.645

callisto · Answer

I'm not sure. But I think mine is better :P (Just kidding)

anonymous · Answer

$$this --\sigma-- stands~ for~ standard ~deviation~ though~righ.t.?$$

callisto · Answer

yes, σ = SD

anonymous · Answer

Ok. Thanks for helping... but I think I'll stick with my interpretation :p :D

callisto · Answer

attachment

anonymous · Answer

Thanks for the pic aswell!

callisto · Answer

Hmm, the pic is for reference, trying to use it to explain my idea. You're welcome

anonymous · Answer

I found the answer, and I made a little mistake. It's 1.645 instead of 0.645, so the answer is 7.29