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? (d) Find a 90% confidence upper bound for the mean thickness.

Step-by-step explanation:

From the information given,

Mean, μ = 4.2 mm

Standard deviation, σ = 0.1 mm

number of sample, n = 87

1) For a confidence level of 98%, the corresponding z value is 2.33.

We will apply the formula

Confidence interval

= mean ± z ×standard deviation/√n

It becomes

4.2 ± 2.33 × 0.1/√87

= 4.2 ± 2.33 × 0.0107

= 4.2 ± 0.025

b) The lower end of the confidence interval is 4.2 - 0.025 =4.18

The upper end of the confidence interval is 4.2 + 0.025 =4.23

c) 0.015 = 2.33 × 0.1/√n

0.015/2.33 = 0.1/√n

0.00644 = 0.1/√n

√n = 0.1/0.00644 = 16

n = 16² = 256

d) For a confidence level of 90%, the corresponding z value is 1.645.

It becomes

4.2 ± 1.645 × 0.1/√87

= 4.2 ± 1.645 × 0.0107

= 4.2 ± 0.0176

The upper bound for the mean thickness is

4.2 + 0.0176 = 4.2176mm

or