If the standard deviation of shear strength measurements for spot welds is 10.5 pounds per square inch (psi), what is the minimum number of test welds that should be sampled if we want the sample mean to be within 0.9 psi of the true mean with probability approximately 0.99? (round your answer up to the nearest whole number.)

The minimum number is 1456.

Step-by-step explanation:

The sheer Stength for spot welds is a random variable with unknown mean μ and standard deviation σ = 10.5. Let X be the sample mean of a sample size of n spot welds. The Central Limit Theorem states that X has approrximately Normal Distribution with mean μ and standard deviation 10.5/(√n). Lets denote with W the standarization of X. W has distribution approximately Normal with mean 0 and standard deviation 1. W is given by the following formula

The cummulative distribution function of W, noted by Ф, has its values tabulated, and they can be found on the attached file. We want a confidence interval of 99%, so we should find Z such that P(-Z < W < Z) = 0.99. Since the Normal density function is symmetric, this is equivalent to find Z such that P(W < Z) = 0.995, in other words, Ф(Z) = 0.995

If we watch the table, we will find that the value of Z corresponding to the value 0.995 is 3.27. This means that

This is equivalent to

Now, we take out the X and the sign, reverting the inequalities, and we obtain

Thus, a 99% confidence interval for μ is .

In order for the sample mean to be within 0.9 psi from the true mean, we need n such that

We take n = 1456 in order to assure that the sample mean is within 0.9 psi of its true mean.

55? i think maybe

step-by-step explanation:

answer:   x = 3 + √11     and     x = 3 - √11

step-by-step explanation:

x² - 6x + 9 = 11

(x - 3)² = 11

x - 3 = ±√11

    x = 3 ± √11

x = 3 + √11     and     x = 3 - √11