[7 points] A sample of 78 anchor bolts was randomly selected and the shear strength (in kips) of each was measured, resulting in a sample mean of 4.25 and a sample standard deviation of 1.30. (a) Calculate a 98% confidence interval to estimate the true mean shear strength. [4 points] (b) Suppose that the shear strength for all bolts in the population was not normally distributed. Would the confidence interval found in part (a) still be valid

Step-by-step explanation:

We want to determine a 98% confidence interval to estimate the true mean shear strength of anchor bolts.

Number of sample, n = 78

Mean, u = 4.25

Standard deviation, s = 1.3

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

4.25 ± 2.33 × 1.3/√78

= 4.25 ± 2.33 × 0.147

= 4.25 ± 0.34

The lower end of the confidence interval is 4.25 - 0.34 = 3.91

The upper end of the confidence interval is 4.25 + 0.34 = 4.59

b) the confidence interval would still be valid because the number of samples is large(> 30)

step-by-step explanation:

(b) 240

step-by-step explanation:

first, we use the pythagorean formula using the given side lengths to find x. then we use x to find the side lengths. then we add the side lengths to find the perimeter.

pythagorean theorem formula:

a^2 + b^2 = c^2

(x - 20)^2 + (x - 40)^2 = x^2

x^2 - 40x + 400 + x^2 - 80x + 1600 = x^2

x^2 -120x + 2000 = 0

(x - 100)(x - 20) = 0

x - 100 = 0 or x - 20 = 0

x = 100 or x = 20

we see that the solution x = 20 must be discarded because it will give a negative side length and a side length of 0. the only valid solution is x = 100.

perimeter = sum of the lengths of the sides

perimeter = x + x - 20 + x - 40

perimeter = 3x - 60

replace x with 100.

perimeter = 3(100) - 60

perimeter = 300 - 60

perimeter = 240