Question

5. -12 points DevoreStat9 5.E.053. My Notes Ask Your Teacher Rockwell hardness of pins of a certain type is known to have a mean value of 50 and a standard deviation of 1.5. (Round your answers to four decimal places.) (a) If the distribution is normal, what is the probability that the sample mean hardness for a random sample of 9 pins is at least 51? (b) What is the (approximate) probability that the sample mean hardness for a random sample of 43 pins is at least 51? You may need to use the appropriate table in the Appendix of Tables to answer this question. Need Help? Read It Talk to a Tutor

Answer 1

bie - Rockwell hareness of pins of a centus Type is known I to have a mean (u) = so and standard deviation (6) 21.5. Assume that the distribution is normal. @ A random sample of 9 pins are selected ie (n)=9. Here sample mean i s also normally distributed with mean (1x) = =50 and standard deviation (6x) = 6 = 1.5 0.50 vo V9 Here we have to use standard normal variate 2 with mean oi and Standard deviation 11 to find below! probability Then to find, LP( Sample mean hardness is at least si) = ? ie. P(x2,51) = 1-P(X551) = 1- P(X-ME I SI-ut) or 68 =1-P/x-ul SI-so 0.50 / = I-P (2 < 2.00 - use 2-table = 1-0.9772 JP (X751) = 0.0228 / 6 A random sample of 43 pins are selected ie. n=43 here, sample mean (a) is also normally approximately distributed with

en (x) = - so and standard deviation (6x) = 6 n = lis 43 = 0,2287 standard normal variate 2 with deviation I to find below Here, we have to use mean i and standard probability then to find, P (Sample mean hardness is at least si) = ? ie. P(X751) = I-P(Assi) = 1-PCX-ux lp you Slaux 68 share - 1-P Xu8 51-50 0,2287. =I-P (2<4.3725 - 1-P (2 <4: - use 2-table Here 315 value of a score = 4.37, then table value then we have to use this probability is is greater than equal to 1. :: P(x>51) = 1 P (2<4.37) 1. P(x751) = 0.0000 /