Question

At a retail​ store, the spending amount per transaction follows normal distribution with mu=50 and sigma=4​,and given you select a sample of 100​ transactions.

a. What is the standard error or the standard deviation for the sample​ mean?

b. What is the probability that sample mean (xbar) is less than $49?

c What is the probability that sample mean (xbar)is between $49 and $50.5?

d. What is the probability that sample mean (xbar) is above $50.1?

e. There is a 35% chance that sample mean (xbar) is above what value?

Answer 1

a)

standard error = (SD/root(N)) = 0.4

b)

P(Y < 49) = P(Y - mean < 49 - mean)
                  = P( (Y - mean)/(SD/root(N)) < (49 - mean)/(SD/root(N))
                  = P(Z < (49 - mean)/(SD/root(N)))
                  = P(Z < (49 - 50)/0.4)
                  = P(Z < -2.5)
                  = 0.006

c)

P(49 < Y < 50.5) = P(49 - mean < Y - mean < 50.5 - mean)
                  = P((49 - mean)/(SD/root(N)) < (Y - mean)/(SD/root(N)) < (50.5 - mean)/(SD/root(N)))
                  = P((49 - mean)/(SD/root(N)) < Z < (50.5 - mean)/(SD/root(N)))
                  = P((49 - 50)/0.4< Z < (50.5 - 50)/0.4)
                  = P(-2.5 < Z < 1.25)
                  = P(Z < 1.25) - P(Z <-2.5)
                  = 0.888

d)

P(Y > 50.1) = P(Y - mean > 50.1 - mean)
                  = P( (Y - mean)/(SD/root(N)) > (50.1 - mean)/(SD/root(N))
                  = P(Z > (50.1 - mean)/(SD/root(N)))
                  = P(Z > (50.1 - 50)/0.4)
                  = P(Z > 0.25)
                  = 1 - P(Z <= 0.25)
                  = 0.401

e)

P(Z>zp)=0.35

P(Z<zp)=0.65

ANS=51.541