Answer:
(a) σx = σ/√n = 7/√60 = 0.90
The sampling distribution of x is approximately normal with mean μx = 48 and standard error σx = $0.90
It is necessary to assume that x has an approximately normal distribution.
(b) x1-bar = 46 and x2-bar = 50
z1 = (x1-bar - μ)/σx = ((46 - 48)/0.90 = -2.2131 and z2 = (x2-bar - μ)/σx = (50 - 48)/0.90 = 2.2131
P(46 < x-bar < 50) = P(-2.2131 < z < 2.2131) = 0.9731
(c) x1 = 46 and x2 = 50
z1 = (x1 - μ)/σ = (46 - 50)/7 = -0.2857 and z2 = (x2 - μ)/σ = (50 - 48)/7 = 0.2857
P(46 < x < 50) = P(-0.2857 < z < 0.2857) = 0.2249
(d) The x-bar distribution is approximately normal while the x distribution is not normal.
(e) The central limit theorem tells us that the standard deviation of the sample mean is much smaller than the population standard deviation. Thus, the average customer is more predictable than the individual customer.
Step-by-step explanation:
(a) σx = σ/√n = 7/√60 = 0.90
The sampling distribution of x is approximately normal with mean μx = 48 and standard error σx = $0.90
It is necessary to assume that x has an approximately normal distribution.
(b) x1-bar = 46 and x2-bar = 50
z1 = (x1-bar - μ)/σx = ((46 - 48)/0.90 = -2.2131 and z2 = (x2-bar - μ)/σx = (50 - 48)/0.90 = 2.2131
P(46 < x-bar < 50) = P(-2.2131 < z < 2.2131) = 0.9731
(c) x1 = 46 and x2 = 50
z1 = (x1 - μ)/σ = (46 - 50)/7 = -0.2857 and z2 = (x2 - μ)/σ = (50 - 48)/7 = 0.2857
P(46 < x < 50) = P(-0.2857 < z < 0.2857) = 0.2249
(d) The x-bar distribution is approximately normal while the x distribution is not normal.
(e) The central limit theorem tells us that the standard deviation of the sample mean is much smaller than the population standard deviation. Thus, the average customer is more predictable than the individual customer.
