Question

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?

Answer 1

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

Answer 2

The sample size n for which  P(19.9 ≤ Y ≤ 20.1) = 0.95 for the considered random variable is 3600 approximately.

What is empirical rule?

According to the empirical rule, also known as 68-95-99.7 rule, the percentage of values of a normally distributed random variable lying within an interval with 68%, 95% and 99.7% of the values lies within one, two or three standard deviations of the mean of the distribution.

$P(\mu - \sigma < X < \mu + \sigma) = 68\%\\P(\mu - 2\sigma < X < \mu + 2\sigma) = 95\%\\P(\mu - 3\sigma < X < \mu + 3\sigma) = 99.7\%$

where we had  where mean of distribution of X is $\mu$standard deviation from mean of distribution of X is $\sigma$ (all percentages are approximations).

For the considered case, we're given that:

$X \sim N(\mu = 20, \sigma = \sqrt{9} = 3)$ (as standard deviation is positive sq. root of variance).

Since we're working with sample with size n, if we take Y = values of each observation from n sized sample, we get:

$\overline{x} = \mu, \\s = \sigma/\sqrt{n}$

Actually, the question is a bit wrong technically, and it had to be $P(19.9 \leq Y \leq 20.1) = 0.95$ because we're taking about the sample and X represents the population and not the sample.

Now, the distribution that Y will pertain is:

$Y \sim N(\overline{x}. s) = N(\mu = 20, s = \sigma/\sqrt{n} = 3/\sqrt{n})$

We need to find the value of n for which

$P(19.9 \leq Y \leq 20.1) = 0.95$

Converting Y to Z (standard normal distribution), we get the needed probability as:

$P(19.9 \leq Y \leq 20.1) = 0.95$

Since the mean of Y is 20, by using second empirical rule:

$P(20 - 2s < Y < 20 +2s) = 95\% = 0.95$

From comparison to $P(19.9 \leq Y \leq 20.1) = 0.95$, we get:

2s = 0.1

s = 0.1/2 = 0.05

Thus, we get:

$s = \dfrac{\sigma}{\sqrt{n}}\\\\0.05 = \dfrac{3}{\sqrt{n}}\\\\n = (\dfrac{3}{0.05})^2=3600$

Thus, the sample size n for which  P(19.9 ≤ Y ≤ 20.1) = 0.95 for the considered random variable is 3600 approximately.

Learn more about standard normal distribution here:

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