Worked examples
The question
Question: samples of 36 are taken from numbers with mean 50 and SD 12: find the mean and the SD of the sample means.
μ = 50, σ = 12, n = 36: mean and SD of x̄?
The central limit theorem: the means of many samples make a bell.
many samples of 36: their means make a bell
1 The center
Its middle is the same as the numbers' mean: 50.
μx̄ = μ = 50
2 The spread
Means of samples are closer together than single numbers.
Their SD is σ divided by the square root of n.
means are less spread out: SD = σ ÷ √n
The square root of 36 is 6.
12 divided by 6 is 2.
σx̄ = σ√n = 12√36 = 126 = 2
Careful!
We divide by the square root of 36, not by 36.
Careful: divide by √n, not by n
The bell of the sample means: middle 50, one SD is 2.
3 Check
Let's check: times the square root of 36 takes us back.
2 times 6 is 12, the SD σ.
So it is right!
2 × √36 = 2 × 6 = 12
Let's look back.
First: the center.
Then: the spread.
Last: check.
Answer Careful: divide by √n, not by n
▶ Watch it solved
The question
Question: samples of 16 are taken from numbers with mean 70 and SD 8: what is the probability that the sample mean is more than 73?
P(x̄ > 73): μ = 70, σ = 8, n = 16
x̄, x bar, is the mean of one sample.
x̄ = the mean of one sample
1 The sample means
The central limit theorem: the means of many samples make a bell.
Its middle is the mean, 70.
Its SD is smaller: σ divided by the square root of n.
many samples of 16: their means make a bell
middle: μ = 70
SD of the means: σ ÷ √n
The square root of 16 is 4.
8 divided by 4 is 2.
σx̄ = σ√n = 8√16 = 84 = 2
Careful!
The sample means have SD 2, not 8.
So we divide by 2.
Careful: not z = 73 − 708 ✗
2 Find z
z counts how many SDs the sample mean is from the mean: x̄ minus the mean, over the SD.
z = x̄ − μσx̄
73 minus 70 is 3.
Divided by 2, that is 1.5.
z = 73 − 702 = 32 = 1.5
3 Draw and shade
The bell: the mean, 70, in the middle.
73 is right of it.
We want the sample mean more than 73.
We shade that part.
4 Read the table
The z-table gives the area to the left of z.
Row 1.5, column .00.
For z = 1.5: the row 1.5 and the column .00 meet at 0.9332.
5 The area we want
The whole bell is 1.
The right side is 1 minus the left: 1 minus 0.9332 is 0.0668.
P(x̄ > 73) = 1 − 0.9332 = 0.0668
6 Check
Let's check another way.
The bell is the same on both sides.
So the area right of 1.5 is the area left of −1.5.
right of 1.5 = left of −1.5
The table at −1.5 gives 0.0668.
The same!
So it is right!
table at −1.5: 0.0668
Let's look back.
First: the sample means.
Then: find z.
Then: draw and shade.
Then: read the table.
Then: the area we want.
Last: check.
Answer P(x̄ > 73) = 1 − 0.9332 = 0.0668
▶ Watch it solved
The question
Question: samples of 25 are taken from numbers with mean 64 and SD 10: what is the probability that the sample mean is between 60 and 66?
P(60 < x̄ < 66): μ = 64, σ = 10, n = 25
x̄, x bar, is the mean of one sample.
x̄ = the mean of one sample
1 The sample means
The central limit theorem: the means of many samples make a bell.
Its middle is the mean, 64.
Its SD is smaller: σ divided by the square root of n.
many samples of 25: their means make a bell
middle: μ = 64
SD of the means: σ ÷ √n
The square root of 25 is 5.
10 divided by 5 is 2.
σx̄ = σ√n = 10√25 = 105 = 2
Careful!
The sample means have SD 2, not 10.
So we divide by 2.
Careful: not z = 60 − 6410 ✗
2 Find z
z counts how many SDs the sample mean is from the mean: x̄ minus the mean, over the SD.
z = x̄ − μσx̄
60 minus 64 is −4.
Divided by 2, that is −2.
z = 60 − 642 = −42 = −2
And for 66: 66 minus 64 is 2.
Divided by 2, that is 1.
z = 66 − 642 = 22 = 1
3 Draw and shade
The bell: the mean, 64, in the middle.
60 is left of it.
66 is right of it.
We want the sample mean between 60 and 66.
We shade that part.
4 Read the table
The z-table gives the area to the left of z.
For z = −2: 0.0228.
For z = 1: 0.8413.
5 The area we want
Between them: the bigger area minus the smaller one.
0.8413 minus 0.0228 is 0.8185.
P(60 < x̄ < 66) = 0.8413 − 0.0228 = 0.8185
6 Check
Let's check: the three parts of the bell.
Left of z = −2: 0.0228.
Right of z = 1: 1 minus 0.8413, 0.1587.
left 0.0228 + middle 0.8185 + right 0.1587
They add up to 1, the whole bell.
So it is right!
= 1.0000
Let's look back.
First: the sample means.
Then: find z.
Then: draw and shade.
Then: read the table.
Then: the area we want.
Last: check.
Answer P(60 < x̄ < 66) = 0.8413 − 0.0228 = 0.8185
▶ Watch it solved
