Assume the distribution is normal. Use the area of the normal curve to answer the question. Round to the nearest whole percent.

Question

The average size of the fish in a lake is 11.4 inches, with a standard deviation of 3.2 inches. Find the probability of catching a fish longer than 17 inches.

I posted another question on standard deviation earlier, but I'm unable to find clear results on how to solve this problem specifically either.

I know the equation for standard deviation but the part I cannot find any information online is how to solve "longer than 17 inches". I keep finding similar question that will ask to find the probability in between two numbers but I can't find much else.

Answer 1

You have $X \sim\mathsf{Norm}(\mu=11.4,\sigma=3.2).$ You seek $P(X > 17) - 1 - P(X \le 17) = 0.0401,$ to four places.

Here is the exact numerical answer from R, where pnorm is a normal CDF.[If you're using a statistical calculator you should be able to use it to get the same answer.]

1 - pnorm(17, 11.4, 3.2)
[1] 0.04005916

Because $17 - 11.4 = 5.6$ and $5.6/3.2 = 1.75$ you're asking for the probability of a fish with length more than 1.75 standard deviations above the mean. By the Empirical Rule the probability in a normal distribution more than $2\sigma$ above the mean is only about 2.5%. So, if you're familiar with this rule, you should not be surprised that the numerical answer is so small.

If you are supposed to get the answer by standardizing and using a printed CDF table of the standard normal distribution, you could begin as follows:

$$P(X \le 17) = P\left(\frac{X=\mu}{\sigma} \le \frac{17 - 11.4}{3.2}\right) = P(Z \le 1.75),$$ where $Z$ has a standard normal distribution.

I will leave it to you to figure out how to get the answer from a printed table. Various textbooks have different styles of normal tables, so you will have to figure out how to use the table in your book. You will start by finding 1.75 in the margins of the table and getting a relevant probability from the body of the table.

In the figure below, the left-hand panel shows the problem on a scale of inches and the right-hand panel shows the 'standardized' version of the problem compatible with printed tables. In both cases the total area under the density curve is $1,$ and the final numerical answer is the area to the right of the dotted vertical line.

R code to make the figure (in case anyone wants to see it):

par(mfrow=c(1,2))
 curve(dnorm(x, 11.4, 3.2), 0, 25, lwd=2, ylab="Density", 
       xlab="Inches", main="NORM(11.4, 3.2)")
  abline(h=0, col="green2")
  abline(v=17, col="red", lwd=2, lty="dotted")
 curve(dnorm(x, 0, 1), -3.5, 3.5, lwd=2, ylab="Density",
       xlab="z", main="NORM(0, 1)")
  abline(h=0, col="green2")
  abline(v=0, col="green2")
  abline(v=1.75, col="red", lwd=2, lty="dotted")
par(mfrow=c(1,1))