Question

Up U29 Details A population of values has a normal distribution with y = 107.5 and o=9.7. A random sample of size T = 133 is drawn. a. Find the probability that a single randomly selected value is between 106.4 and 110. Round your answer to four decimal places. P(106.4 < X <110) = b. Find the probability that a sample of size n = 133 is randomly selected with a mean between 106.4 and 110. Round your answer to four decimal places. P(106.4 < M <110) = Submit Question SAP E YouTube po Pre

Answer 1

Given, u = 107.5 0=9.7 n=133 The required probability is P(1064 < X <110) = P(106,4-4-4-44, 11074) =P(1964–107.5 <z«1101075 9.7 9.7 = P(-0.11<Z <0.26) = P(Z <0.26)-P(Z <-0.11) = 0.6026 -0.4562 = 0.1464 (From the standard normal table)

The required probability is, P(106.4 <M <110)=P 106.4-4 M-u 110 - u (on on a 106.4-107.5 7 110-107.5 9.7/2133 9 .7/ 133 = P(-1.31<Z <2.97) =P(Z <2.97)-P(Z <-1.31) =0.9985 -0.0951 =0.9034 (Using standard normal table)