The breaking strength of a rivet has a mean value of 10,100 psi and a standard deviation of 499 psi. (a) What is the probability that the sample mean breaking strength for a random sample of 40 rivets is between 10,000 and 10,300? (Round your answer to four decimal places.) (b) If the sample size had been 15 rather than 40, could the probability requested in part (a) be calculated from the given information? Explain your reasoning O Yes, the probability in part (a) can still be calculated from the given information O No, n should be greater than 30 in order to apply the Central Limit Theorem. O No, n should be greater than 20 in order to apply the Central Limit Theorem. O No, n should be greater than 50 in order to apply the Central Limit Theorem. You may need to use the appropriate table in the Appendix of Tables to answer this question.

We would assume a normal distribution for the breaking strength of a rivet. We would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ/√n

Where

n = number of samples

x = Breaking strengths of rivet.

µ = mean breaking strength

σ = standard deviation

From the information given,

µ = 10,100 psi

σ = 499 psi

n = 40

The probability that the sample mean breaking strength for a random sample of 40 rivets is between 10000 and 10,300 is expressed as

P(10000 ≤ x ≤ 10300)

For x = 10000,

z = (10000 - 10100)/499/√40 = - 0.13

Looking at the normal distribution table, the probability corresponding to the z score is 0.45

For x = 10300,

z = (10300 - 10100)/499/√40 = 2.54

Looking at the normal distribution table, the probability corresponding to the z score is 0.99

Therefore,

P(10000 ≤ x ≤ 10300) = 0.99 - 0.45 = 0.54

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