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Normal: P(X < 63.5), mean 60, sd 2.5

Step by step, with the reason for every step · Normal probabilities with the z-table · University Statistics

Step by step

The question

Question: X is normal, with mean 60 and SD 2.5: what is the probability that X is less than 63.5?

P(X < 63.5): μ = 60, σ = 2.5

μ is the mean.

σ is the standard deviation, the SD.

μ = the mean, σ = the SD

1 Find z

z counts how many SDs x is from the mean: x minus the mean, over the SD.

z = x − μσ

63.5 minus 60 is 3.5.

Divided by 2.5, that is 1.4.

z = 63.5 − 602.5 = 3.52.5 = 1.4

2 Draw and shade

The bell: the mean, 60, in the middle.

63.5 is right of it.

We want X less than 63.5.

We shade that part.

3 Read the table

The z-table gives the area to the left of z.

Row 1.4, column .00.

For z = 1.4: the row 1.4 and the column .00 meet at 0.9192.

4 The area we want

Careful!

The z-table always gives the area to the left of z.

Here that is what we want.

Careful: the table gives the area to the LEFT

Less than: the area to the left.

That is just the table number, 0.9192.

P(X < 63.5) = 0.9192

5 Check

Let's check another way.

The bell is the same on both sides.

So the area left of 1.4 is the area right of −1.4.

left of 1.4 = right of −1.4

1 minus the table at −1.4: 1 minus 0.0808 is 0.9192.

The same!

So it is right!

1 − table at −1.4: 1 − 0.0808 = 0.9192

Let's look back.

First: find z.

Then: draw and shade.

Then: read the table.

Then: the area we want.

Last: check.

Answer
P(X < 63.5) = 0.9192

Your turn

Try this one yourself, then check:

Find P(X < 60.5): μ = 50, σ = 7.5
Show the answer
P(X < 60.5) = 0.9192

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