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

45%

Step-by-step explanation: