Introduction to statistics

Question

Suppose a population has µ = 8 and δ= 3. Suppose a random sample of n= 36 is selected. What is the probability that the sample mean is between 7.8 and 8.2?

Answer 1

want to find P(7.8 < xbar < 8.2)

transform inequality to standard normal:

P( (xbar LL - mean)/(sd/sqrt(ss)) < z < (xbar UL - mean)/(sd/sqrt(ss))

where xbar LL = 7.8, xbar UL = 8.2, mean = 8, sd = 3

P( (7.8 - 8)/(3/6) < z < (8.2 - 8)/(3/6)) = P( -0.4 < z < 0.4) = P(z < 0.4) - P(z < -0.4) = 0.6554 - 0.3446 = 0.3108