In a sample of 100 steel wires the average breaking strength is 50 kN, with a standard deviation of 2 kN.

a. Find a 95% confidence interval for the mean breaking strength of this type of wire.

b. Find a 99% confidence interval for the mean breaking strength of this type of wire.

c. An engineer claims that the mean breaking strength is between 49.7 kN and 50.3 kN. With what level of confidence can this statement be made?

d. How many wires must be sampled so that a 95% confidence interval specifies the mean breaking strength to within ±0.3 kN?

e. How many wires must be sampled so that a 99% confidence interval specifies the mean breaking strength to within ±0.3 kN?

Step 1 of 5

Given sample size n=100

Mean breaking strength kn

Standard deviation is =2 kn

a) We have to find 95% confidence interval for average breaking strength

The 95% confidence interval is

=

=

=(49.608, 50.392)

Hence the 95% confidence intervals is (49.608, 50.392)

Step 2 of 5</p>

b) We have to find 99% confidence interval for average breaking strength

The 99% confidence interval is

=

=

=(49.484, 50.516)

Hence the 99% confidence intervals is (49.484, 50.516)

Step 3 of 5</p>

c) We have to find the confidence level

The mean value is between 49.7 and 50.3

Then

49.7<<50.3

The lower limit is =49.7

=49.7

(0.2)=50-49.7

=1.5

Find the area to the right of the Z value=1-0.93319

=0.06681

Then =0.06681

=0.13362

1-=1-0.13362

=0.86638

=86.64%

Hence the confidence level is 86.64%