Question

Questions:

Suppose a team of biologists has been studying a fishing pond. Let x represent the length of a single trout taken at
random from the pond. This group of biologists has determined that x (trout length) has a normal distribution with
mean (mu) 10.5 inches and standard deviation (sigma) 1.2 inches. Use calculator command (normalcdf) to find
your answer. Show your work by writing the command and numbers that you input. Round your answer to 4
decimal places. Circle your final answers.
1. What is the probability that the length of a single trout taken at random from the pond is between 7 and 11
inches long?
2. What is the probability that the mean length of five trout taken at random is between 7 and 11 inches?

Answer 1

1. To find the probability that the length of a single trout taken at random from the pond is between 7 and 11 inches long, we need to use the normalcdf command with the given parameters. The formula for normalcdf is normalcdf(lower bound, upper bound, mean, standard deviation).

Thus, normalcdf(7, 11, 10.5, 1.2) = 0.6827.

Therefore, the probability that the length of a single trout taken at random from the pond is between 7 and 11 inches long is 0.6827.

2. To find the probability that the mean length of five trout taken at random is between 7 and 11 inches, we need to use the central limit theorem, which states that the sample mean will also follow a normal distribution with a mean of the population mean and a standard deviation of the population standard deviation divided by the square root of the sample size (i.e., sigma/sqrt(n)).

Thus, we first need to find the standard deviation of the sample mean, which is sigma/sqrt(n) = 1.2/sqrt(5) = 0.5367.

Then, we can use the normalcdf command with the given parameters. The formula for normalcdf is normalcdf(lower bound, upper bound, mean, standard deviation).

Thus, normalcdf(7, 11, 10.5, 0.5367) = 0.9468.

Therefore, the probability that the mean length of five trout taken at random is between 7 and 11 inches is 0.9468.