# Thread: Normal distribution question involving samples

1. ## Normal distribution question involving samples

Q: The mass of coffee in a randomly chosen jar sold by a certain company may be taken to have a normal distribution with mean 203g and standard deviation 2.5g. The random variable C denotes the mean mass (in grams) of coffee jar per in a random sample of 20 jars, find the value of a such that P(|C - 203| < a) = 0.95.

How to solve this question?

2. Originally Posted by Cathelyn13
Q: The mass of coffee in a randomly chosen jar sold by a certain company may be taken to have a normal distribution with mean 203g and standard deviation 2.5g. The random variable C denotes the mean mass (in grams) of coffee jar per in a random sample of 20 jars, find the value of a such that P(|C - 203| < a) = 0.95.

How to solve this question?

You should realise that $C$ ~ Normal $\left( \mu = 203, ~ \sigma = \frac{2.5}{\sqrt{20}}\right)$.

Therefore $X = C - 203$ ~ Normal $\left( \mu = 0, ~ \sigma = \frac{2.5}{\sqrt{20}}\right)$.

Now note that $\Pr(|X| < a) = \Pr(-a < X < a) = 1 - 2\Pr (X > a)$ by symmetry.

3. I see. But why the standard deviation has to be divided by the root of the sample size? Is there any formula to the finding the sample mean and sample standard deviation from the population mean and standard deviation?

4. Originally Posted by Cathelyn13
I see. But why the standard deviation has to be divided by the root of the sample size? Is there any formula to the finding the sample mean and sample standard deviation from the population mean and standard deviation?
Have you covered the sampling distribution of the mean in class? The formula I used will be in your class notes or textbook.

5. I think so . Are you referring to: If X ~ N ( μ, δ ²) then X bar ~ N (μ, δ ²/n)? Using this I can get the answer now. Thanks!!! =)