Question

using the normal distribution of fish lengths for which mean is 9 incehs and the standard deviation is 2 inches. Assume the variable x is normal distribution. A:) What percent of fish are longer than 11"? B:) If 400 fish are randomly selected, about how many would you expect to be shorter than 7"?

Answer 1

Let \(X\) denote a random variable for length of a fish. If \(Z\) denotes a random variable with a standard normal distribution
\[P(X>11)=P\left(Z>\frac{11-9}{2}\right)=P(Z>1)\approx0.1587\]
About 15.87% of fish are longer than 11''.

You use a similar set up for the next part, with some slight changes:
\[P(X<7)=P\left(Z<\frac{7-9}{2}\right)=P(Z<-1)\approx0.1587\]
If 400 fish are randomly selected, you can expect somewhere around 15.87% of 400, or 63.48 fish. Round up to 64.