The breaking strength of standard deviation of 503 psi. a rivet has a mean value

Question

The breaking strength of standard deviation of 503 psi. a rivet has a mean value of 10,000 psi and a (a) What is the probability that the sample mean breaking strength for a random sample of 40 rivets is between 9,900 and 10,200? (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 O Yes, the probability in part (a) can still be calculated from the given O No, n should be greater than 30 in order to apply the Central Limit O No, n should be greater than 20 in order to apply the Central Limit O No, n should be greater than 50 in order to apply the Central Limit your reasoning information. Theorem. Theorem Theorem. You may need to use the appropriate table in the Appendix of Tables to answer this question.

Explanation / Answer

Solution:- = 10000 , = 503

Formula Z = (X - )/(/sqrtn)

a) n = 40, probability between 9900 and 10200

P(9900 < X < 10200) = P((9900-10000)/(503/sqrt40) < Z < (10200-10000)/(503/sqrt40)

= P(-1.2574 < Z < 2.5147)

   =  0.8902

b) n = 15

  P(9900 < X < 10200) = P((9900-10000)/(503/sqrt15) < Z < (10200-10000)/(503/sqrt15)

= P(0.7700 < Z < 1.5400)

=  0.1588

=> option B. No,n should be greater than 30 in order to apply the central limit theorem.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.