Question

Question 5. A lahmacun company has found that service times for orders follow a normal distribution with a mean of 40 minutes and a standard deviation of 10 minutes. A random sample of five orders was taken. a) The probability is 0.2 that the sample mean of service times is less than how many minutes? (10 pts) b) What is the probability that more than one of these orders take more than 60 minutes? (10 pts)

Answer 1

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953 Giren, mean = Mi 40, SD =0=10., n=5. Sample mean = x Ñ ~ N C Mino) where poate=4147 By (intral limit theorem, Z=X - 40-N108) м- 4o, 4.41 a) det consider time a minutes Parza) ; 0.2 -> pl X-40 a ho ho -0.2 Play = 0.2 => $410.2) :-0.842 a 40 < )-0.2.-) PLZ < 4.47 4.47 -) a-40 a-40 4.47 -) U: 31:24. So, required times 36:24 minutes by Roxxx tt b> 0 Here X~ N(40, 10), BY CLY, Z: X-40 ~ N(014) P1 * 760) = f( 1940) = {(272) - 1-P (252) : 1 - 0.9772 0.0228 f (x710)=0.0228 X-40 10 > 60