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...

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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.