Question

A population of values has a normal distribution with u = a random sample of size n = 16. 229.4 and o = 67.4. You intend to draw Find the probability that a single randomly selected value is greater than 212.6. P(X > 212.6) = Find the probability that a sample of size n = 16 is randomly selected with a mean greater than 212.6. P(M> 212.6) = Enter your answers as numbers accurate to 4 decimal places. Answers obtained using exact 2-scores or 2-scores rounded to 3 decimal places are accepted. 2 Calculator

Answer 1

Given : M= 229.4, T-67.4 16 a) p( single randomly selected value as greater than 212.6) PCX> 212.6) = P(X=4>212.6– 229.4. 67.4 =P(Z > -0.249258.--)) P(Z>-0.2493) normaleaf (-0,2493, ,1099,0,0) 0.5984355833 (From T1-84) + P(x > 212.6) = 0.5984 Answer) b) p(a sample size of n=16 as randomly selected mean greater than 212.6) ·P(M>212.6) = with a 2120.6-229.4 = EP (M-u 67.4/516 =P(Z > -0.997032 64 --) - P(Z >-0.9970) = normal edf (-0.9970,1099, 0, 0 0.8406177406.. ( From T1-84) :: P(*>212.6) - (Answer) 0.8406