# The lengths of lumber a machine cuts are normally distributed with a mean of 90 inches and a standard deviation of 0.3 inch.(a) What is the probability that a randomly selected board cut by the machine has a length greater than 90.09 inches?(b) A sample of 42 boards is randomly selected. What is the probability that their mean length is greater than 90.09 inches?

(a) The probability of board being longer than 90.09 is about 38%.

(b) The probability of a sample (n=42) having a mean length of 90.09 is about 2.6% We are given a distribution of board lengths and told that they are approximately normal with a mean `mu=90` and a population standard deviation of `sigma=0.3` .

(a) We are asked to find the probability that a randomly chosen board has length greater than90.09 `P(x>90.09)`

We can convert the length to a normal value and use a standard normal chart to get the probability. (Note that most standard normal charts give the probability that a chosen value is less than the given value so we would use the complement.)

`z=(x-mu)/sigma=(90.09-90)/.3=0.30`

From a chart (or utility) we find `P(z>.3)~~.3821`

(b) We are asked to find the probability that a sample size n=42 has a mean `bar(x)=90.09` . Here we apply the central limit theorem. The central limit theorem essentially says that for a given sample size if we were able to list all of the possible samples of that size this distribution would be normal with mean `mu` and standard deviation `sigma/sqrt(n)` which is called the standard error of the mean.

Here we get `z=(90.09-90)/(.3/sqrt(42))~~1.94` and `P(z>1.94)~~.0261`

Notice the disparity between finding an individual with a given characteristic and a group with the same characteristic. This is the reason we can calculate statistics for entire populations from relatively small samples.