Step by step
The question
Question: samples of 36 are taken from numbers with mean 30 and SD 9: what is the probability that the sample mean is less than 32?
x̄, x bar, is the mean of one sample.
There are four choices.
First we solve it, the normal way.
Then we pick.
1 The sample means
The central limit theorem: the means of many samples make a bell.
Its middle is the mean, 30.
Its SD is smaller: σ divided by the square root of n, where n is the sample size.
In a mean, high and low numbers cancel out part of the way.
That is why the spread shrinks.
The square root of 36 is 6.
9 divided by 6 is 1.5.
Careful!
The sample means have SD 1.5, not 9.
So we divide by 1.5.
2 Find z
z counts how many SDs the sample mean is from the mean: x̄ minus the mean, over the SD.
Every normal bell has the same shape when we count in SDs.
So one table, the z-table, works for every bell.
32 minus 30 is 2.
Divided by 1.5, that is about 1.33.
3 Draw and shade
The bell: the mean, 30, in the middle.
32 is right of it.
We want the sample mean less than 32.
We shade that part.
The whole bell has area 1, and the shaded area is the probability.
4 Read the table
The z-table gives the area to the left of z.
Row 1.3, column .03.
For z = 1.33: the row 1.3 and the column .03 meet at 0.9082.
5 The area we want
Less than: the area to the left.
That is just the table number, 0.9082.
6 Check
Let's check another way.
The bell is the same on both sides.
So the area left of 1.33 is the area right of −1.33.
1 minus the table at −1.33: 1 minus 0.0918 is 0.9082.
The same!
So it is right!
7 Pick the choice
Our answer is P(x̄ < 32) = 0.9082.
That is choice D.
8 Why not the others
A is 0.5879.
Wrong: used the SD 9: the sample mean's SD is 9/√36 = 1.5.
B is 0.9772.
Wrong: used z = 2: z = 2/1.5 ≈ 1.33.
C is 0.0918.
Wrong: that is P(more than 32).
Let's look back.
First: the sample means.
Then: find z.
Then: draw and shade.
Then: read the table.
Then: the area we want.
Then: check.
Then: pick the choice.
Last: why not the others.