Normal: P(X < 60), mean 50, sd 10 Easy

The question

Question: X is normal, with mean 50 and SD 10: what is the probability that X is less than 60?

μ is the mean.

σ is the standard deviation, the SD.

1 Find z

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

60 minus 50 is 10.

Divided by 10, that is 1.

2 Draw and shade

The bell: the mean, 50, in the middle.

60 is right of it.

We want X less than 60.

We shade that part.

3 Read the table

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

Row 1.0, column .00.

For z = 1: the row 1.0 and the column .00 meet at 0.8413.

4 The area we want

Careful!

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

Here that is what we want.

Less than: the area to the left.

That is just the table number, 0.8413.

5 Check

Let's check another way.

The bell is the same on both sides.

So the area left of 1 is the area right of −1.

1 minus the table at −1: 1 minus 0.1587 is 0.8413.

The same!

So it is right!

Let's look back.

First: find z.

Then: draw and shade.

Then: read the table.

Then: the area we want.

Last: check.

▶ Watch it solved

Normal: P(45 < X < 60), mean 50, sd 10 Medium

The question

Question: X is normal, with mean 50 and SD 10: what is the probability that X is between 45 and 60?

μ is the mean.

σ is the standard deviation, the SD.

1 Find z

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

45 minus 50 is −5.

Divided by 10, that is −0.5.

And for 60: 60 minus 50 is 10.

Divided by 10, that is 1.

2 Draw and shade

The bell: the mean, 50, in the middle.

45 is left of it.

60 is right of it.

We want X between 45 and 60.

We shade that part.

3 Read the table

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

For z = −0.5: 0.3085.

For z = 1: 0.8413.

4 The area we want

Careful!

Both table numbers start at the far left.

So we take away the smaller one.

Between them: the bigger area minus the smaller one.

0.8413 minus 0.3085 is 0.5328.

5 Check

Let's check: the three parts of the bell.

Left of z = −0.5: 0.3085.

Right of z = 1: 1 minus 0.8413, 0.1587.

They add up to 1, the whole bell.

So it is right!

Let's look back.

First: find z.

Then: draw and shade.

Then: read the table.

Then: the area we want.

Last: check.

▶ Watch it solved

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

The question

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

μ is the mean.

σ is the standard deviation, the SD.

1 Find z

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

63.5 minus 60 is 3.5.

Divided by 2.5, that is 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.

Less than: the area to the left.

That is just the table number, 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.

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

The same!

So it is right!

Let's look back.

First: find z.

Then: draw and shade.

Then: read the table.

Then: the area we want.

Last: check.

▶ Watch it solved