To solve this problem, we can use the z statistic to find
for the probability. The formula for z score is:
z = (x – u) / s
Where,
u = the sample mean = 68
s = standard deviation of the samples = 12
x = sample value = 70
In this case, what is asked is the probability that the
sample mean is larger than 70, therefore this corresponds to the right of z on
the graph of normal distribution.
z = (70 – 68) / 12
z = 2 / 12
z = 0.17
Therefore finding for P at z ≥ 0.17 using the standard
distribution tables:
P (z = 0.17) = 0.5675
But this is not the answer yet since this is the P to the
left of z. Therefore the correct one is:
P (z ≥ 0.17) = 1 - 0.5675
P (z ≥ 0.17) = 0.4325 = 43.25%
<span>The probability that the sample mean is larger than 70 is
43.25%</span>
