Samples of 25 from mean 64, sd 10: P(60 < sample mean

Step by step

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.

σ_x̄ = 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 − μσ_x̄

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