A sample of 87 glass sheets has a mean thickness of 4.20 mm with a standard deviation of 0.10 mm.

a. Find a 98% confidence interval for the population mean thickness.

b. What is the level of the confidence interval (4.185, 4.215)?

c. How many glass sheets must be sampled so that a 98% confidence interval will specify the mean to within ±0.015?

Step 1 of 3:

Let n denotes the sample of 87 glass.

Let denotes the mean thickness is 4.20 mm and

Let s denotes the standard deviation is 0.10 mm.

From the given information we know that n=87, =4.20 mm and s=0.10 mm.

Our goal is:

a). We need to find the 98% confidence interval for the population mean thickness.

b). We need to find the level of the confidence interval (4.185,4.215).

c). We need to find how many glass sheets must be sampled so that a 98% confidence

interval will specify the mean to within 0.015.

a).

Now we need to find the 98% confidence interval for the population mean thickness.

The formula of the confidence interval is

The 98% of confidence interval of is

2.33.

Now we have to find the confidence interval.

We know that ,z and n values.

Substitute all the values.

Therefore, the confidence interval is 4.2249 and 4.1751.

Step 2 of 3:

b).

Now we have to find the level of the confidence interval (4.185,4.215).

4.215=

4.215=

.

Then the area of the right angle z is 1-1.40

From the area under the normal curve.

=1-0.9192

=0.0808=.

Then,1-.

=1-2(0.0808)

=1-(0.1616)

=0.8384.

Therefore, the level of the confidence interval is 83.84%.