Question

Question 6 2 pts An unknown distribution has a mean of 80 and a standard deviation of 13. Samples of size n-35 are drawn randomly from the population. Find the probability that the sample mean is between 82 and 92. (round to 4 decimal places) Example page 397 Wk6Hw_SmpMean 1

Answer 1

Given Input:

Sample Mean (X) = 80 , std deviation (sd) = 13, n = 35 Y = 82, Y* = 92

Formula for Z- Score Z = (Y- X)/ (sd/sqrt(n))

Z( 82) = (80-82) / 13 * sqrt (35)= -0.91

Using Z Score table

P( Y< 82) = 0.81835

P (Y>82) = 1 - P(Y<82) = 0.18165

P (80 < Y < 82) = P (Y < 82) -0.5 = 0.31835

Z( 92) = (92- 80) / 13 * sqrt (35)= 5.4545

Using Z Score table

P( Y< 92) = 1

P (Y>92) = 1 - P(Y<92) = 0

P (80 < Y < 92) = P (Y < 92) -0.5 = 0.5

P (82< Y < 92) = P ( Y <92) - P (Y> 82)

P (82< Y < 92) = (1 - 18 165) = .8183 ~ 81. 83%