population givens:
μ=12, σ=1.5
sample:
μs = μp (mean of the sample mean is the same as the mean of the population)
σs = σp/√n (std dev of the sample mean equals the popul std dev divided by the sqrt of the sample size)
z = (x - μ) / (σ/√n) (formula needed when the sample size is given)
(a) Find the mean and standard deviation of the sample mean of randomly chosen
samples of size n = 90.
μp = μs = 12
σs = 1.5/√90 = 0.1581
(b) Find the probability that the sample mean x of a randomly chosen of size n = 90
is less than 11.7.
z = (x - μ) / (σ/√n)
z = (11.7 - 12) / (1.5/√90) = -1.88
[on z table] p(x < 11.7) = 0.0301 (probability to the left of x)
(c) Find the probability that the sample mean x of a randomly chosen of size n = 90
is greater than 12.3.
z = (12.3 - 12) / (1.5/√90) = 1.90
[on z table] p(x > 12.3) = 1 - 0.9713 = 0.0287 (area to the right equals1 minus the area to the left of x)
(d) Find the probability that the sample mean x of a randomly chosen of size n = 90
is less than 11.7 or greater than 12.3.
p(x < 11.7 or x > 12.3) = 0.0301 + 0.0287 = 0.0588 (add the areas)
