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Normal: P(X > 71.2), mean 65, sd 4

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 65 and SD 4: what is the probability that X is more than 71.2?

P(X > 71.2): μ = 65, σ = 4

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

71.2 minus 65 is 6.2.

Divided by 4, that is 1.55.

z = 71.2 − 654 = 6.24 = 1.55

2 Draw and shade

The bell: the mean, 65, in the middle.

71.2 is right of it.

We want X more than 71.2.

We shade that part.

3 Read the table

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

Row 1.5, column .05.

For z = 1.55: the row 1.5 and the column .05 meet at 0.9394.

4 The area we want

Careful!

0.9394 is the area to the left.

We want the right side.

Careful: not 0.9394 ✗

The whole bell is 1.

The right side is 1 minus the left: 1 minus 0.9394 is 0.0606.

P(X > 71.2) = 1 − 0.9394 = 0.0606

5 Check

Let's check another way.

The bell is the same on both sides.

So the area right of 1.55 is the area left of −1.55.

right of 1.55 = left of −1.55

The table at −1.55 gives 0.0606.

The same!

So it is right!

table at −1.55: 0.0606

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 > 71.2) = 1 − 0.9394 = 0.0606

Your turn

Try this one yourself, then check:

Find P(X > 81.2): μ = 75, σ = 4
Show the answer
P(X > 81.2) = 0.0606

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