Question

Suppose a team of biologists has been studying the Pinedale children's fishing pond. Let x represent the length of a single trout taken at random from the pond. This group of biologists has determined that the length has a normal distribution with mean of 10.2 inches and standard deviation of 1.4 inches.

A. What is the probability that a single trout taken at random from the pond is between 8.5 and 12 inches long ?

B. What are the mean and standard deviation of the mean length of a trout in the Pinedale children's fishing pond for a simple random sample of 40 trouts.

C. What is the probability that the mean length x of 40 trouts taken at random is between 8.5 and 12 inches?

D. Would your method of answering (a) or (c) be affected if the distribution of the trouts length in the Pinedale children's pond were distinctly nonnormal (for example,if they were skewed?) Explain which parts could be answered the same way, which could not, and how you know.

Answer 1

Solution:

We are given that: x represent the length of a single trout taken at random from the pond. This group of biologists has determined that the length has a normal distribution with mean of 10.2 inches and standard deviation of 1.4 inches.

That is: x ~ Normal(

Part A) What is the probability that a single trout taken at random from the pond is between 8.5 and 12 inches long ?

That is: P( 8.5 < X < 12 ) = ...........?

Find z scores:

Thus

P( 8.5 < X < 12 ) = P( -1.21 < Z < 1.29)

P( 8.5 < X < 12 ) = P( Z < 1.29) - P( Z < -1.21 )

Look in z table for z = 1.2 and 0.09 as well as for z = -1.2 and 0.01 and find area.

From z table we get:

P( Z < 1.29 ) = 0.9015

P( Z < -1.21) = 0.1131

P( 8.5 < X < 12 ) = P( Z < 1.29) - P( Z < -1.21 )

P( 8.5 < X < 12 ) = 0.9015 - 0.1131

P( 8.5 < X < 12 ) = 0.7884

Part B) What are the mean and standard deviation of the mean length of a trout in the Pinedale children's fishing pond for a simple random sample of 40 trouts.

n = sample size = 40

Mean of sample means:

Standard deviation of sample means:

Part C) What is the probability that the mean length x of 40 trouts taken at random is between 8.5 and 12 inches?

That is we have to find:

Find z scores:

Thus we get:

Since z = 8.13 very large above the mean = 0 ( for z mean is 0 and standard deviation is 1)

and -7.68 is very small below mean = 0

Thus

P( Z< 8.13) = 1.0000

and

P( Z < -7.68) =0.0000

Thus

Part D) Would your method of answering (a) or (c) be affected if the distribution of the trouts length in the Pinedale children's pond were distinctly nonnormal (for example,if they were skewed?) Explain which parts could be answered the same way, which could not, and how you know.

if the distribution of the trouts length in the Pinedale children's pond were distinctly nonnormal (for example,if they were skewed , then method of answering part A) would be affected.

Method of answering in part C) wouldn't be affected, since sample size = n = 40 > 30, so we can assume large sample and hence applying central limit theorem , we can assume approximate Normal distribution.