Given a normal random variable X with mean 20 and variance 9, and a random sample of size n taken from the distribution, what sample size n is necessary in order that P (19.9 ≤ X ≤ 20.1) = 0.95?

Interval at 95% = mean + / - Z*sd/Sqrt (n)

It can be noted that: error = 20.1-20=20-19.9 = 0.1

Therefore,

0.1 = Z*sd/sqrt (n) = > n = {Z*sd/0.1}^2

At 95% confidence interval, Z = 1.96, sd = sqrt (variance) = sqrt (9) = 3

Then,

Necessary sample size is,

n = {1.96*3/0.1}^2 = 3457.44 = 3,458