The overhead reach distances of adult females are normally distributed with a mean of 197.5 cm197.5 cm and a standard deviation of 7.8 cm7.8 cm. a. find the probability that an individual distance is greater than 207.50207.50 cm. b. find the probability that the mean for 2020 randomly selected distances is greater than 196.00 cm.196.00 cm. c. why can the normal distribution be used in partâ (b), even though the sample size does not exceedâ 30?

Given:
μ = 197.5 cm, the population mean
σ = 7.98 cm, the population standard deviation

Part a.
For the random variable x = 207.50 cm, the z-score is
z = (x-μ)/σ = (207.5-197.5)/7.98 = 1.253
From standard tables, obtain
P(x < 207.5) = 0.895

0.895

Part b.
n = 20, the sample size
σ/√n = 7.98/√(20) = 1.7844

The random variable is x = 196.0 cm
The z-score is
z = (x-μ)/(σ/√n) = (196 - 197.5)/1.7844 = -0.8406
From standard tables, obtain
P(x < 196) = 0.2003

0.2003

Part c.
In part b, the sample size of 30 is less than the minimum recommended value of 30.
The reasons why the normal distribution can be used are
(i) the z-score takes the sample size into account because σ is replaced by
    σ/√n.
(ii) according to the Central Limit Theorem, sample means are normally
     distributed.