# Normal Probability Calculator for Sampling Distributions

Instructions: This Normal Probability Calculator for Sampling Distributions will compute normal distribution probabilities for sample means $$\bar X$$, using the form below. Please type the population mean ($$\mu$$), population standard deviation ($$\sigma$$), and sample size ($$n$$), and provide details about the event you want to compute the probability for (for the standard normal distribution, the mean is 0 and the standard deviation is 1): Population Mean ($$\mu$$) Population St. Dev. ($$\sigma$$) Sample Size ($$n$$)
Two-Tailed:
$$\le \bar X \le$$
Left-Tailed:
$$\bar X \le$$
Right-Tailed:
$$\bar X \ge$$

## More About this Normal Distribution Probability Calculator for Sampling Distributions Tool

When a sequence of normally distributed variables $$X_1, X_2, ...., X_n$$ is averaged, we get the sample mean

$\bar X = \frac{1}{n}\sum_{i=1}^n X_i$

Since any linear combination of normal variables is also normal, the sample mean $$\bar X$$ is also normally distributed (assuming that each $$X_i$$ is normally distributed). The distribution of $$\bar X$$ is commonly referred as to the sampling distribution of sample means .

### How do you calculate sampling distribution?

Assuming that $$X_i \sim N(\mu, \sigma^2)$$, for all $$i = 1, 2, 3, ...n$$, then $$\bar X$$ is normally distributed with the same common mean $$\mu$$, but with a variance of $$\displaystyle\frac{\sigma^2}{n}$$.

This tells us that $$\bar X$$ is also centered at $$\mu$$ but its dispersion is less than that for each individual $$X_i$$. Indeed, the larger the sample size, the smaller the dispersion of $$\bar X$$.

### The sampling distribution formula

The key when working with sampling distributions is to use the fact that if $$\mu$$ is the mean of the population and $$\sigma$$ is the standard deviation of the population, then

$\displaystyle \frac{\bar X - \mu}{\sigma}$

have a standard normal distribution. This is crucial, because we can use this to reduce all sampling distributions into standard normal probability calculations.

### What is the mean of the sampling distribution

The mean of sampling distributions, $$\mu(\bar X)$$, is the same as the underlying mean of the distribution $$\mu$$.

### Standard deviation of sampling distribution

Unlike the case of the mean, the standard deviation of sample means can be calculated using the formula:

$s(\bar X) = \displaystyle \frac{\sigma}{\sqrt n}$

### Calculators Related to the normal distribution

If you want to compute normal probabilities for one single observation $$X$$, you can use this calculator with $$n=1$$, or you can use our regular normal distribution calculator .

Often times you are interested in the reverse process: Given a probability, you want to find the score such as the probability to the right of that score is that given probability, for which you can use a invnorm calculator

Also, if graphical visualization is what you need, you can try directly our normal distribution graph creator. ### Example:

Question: Consider a normal distribution where the population mean is 12, and the population standard deviation is 3.4. Assume you take samples of size n = 16. What is the probability for the sample means to be in the interval (11.3, 12.4)?

Solution:

The following are the population mean $$(\mu)$$, population standard deviation $$(\sigma)$$ and sample size $$(n)$$ provided:

 Population Mean $$(\mu)$$ = $$12$$ Population Standard Deviation $$(\sigma)$$ = $$3.4$$ Sample Size $$(n)$$ = $$16$$ Event to compute its probability = $$11.3 \leq \bar X \leq 12.4$$

We need to compute $$\Pr(11.3 \leq \bar X \leq 12.4)$$. The corresponding z-values needed to be computed are:

$Z_{lower} = \frac{X_1 - \mu}{\sigma/\sqrt{n}} = \frac{ 11.3 - 12}{ 3.4/\sqrt{16}} = -0.82$ $Z_{upper} = \frac{X_2 - \mu}{\sigma/\sqrt{n}} = \frac{ 12.4 - 12}{ 3.4/\sqrt{16}}= 0.47$

Using the properties of the normal distribution, if $$X ~ N(\mu, \sigma)$$, then the variables $$Z_{lower} = \displaystyle \frac{X_1 - \mu}{\sigma/\sqrt{n}}$$ and $$Z_{upper} = \displaystyle \frac{X_2 - \mu}{\sigma/\sqrt{n}}$$ have a standard normal distribution. Therefore, the probability is computed as:

$\begin{array}{ccl} \Pr(11.3 \leq \bar X \leq 12.4) & = & \Pr\left(\displaystyle \frac{ 11.3 - 12}{ 3.4 / \sqrt{ 16}} \leq \frac{ \bar X - 12}{ 3.4 / \sqrt{ 16}} \leq \frac{ 12.4 - 12}{ 3.4 / \sqrt{ 16}}\right) \\\\ \\\\ & = & \displaystyle\Pr\left(\frac{ 11.3 - 12}{ 3.4 / \sqrt{ 16}} \leq Z \leq \frac{ 12.4 - 12}{ 3.4 / \sqrt{ 16}}\right) \\\\ \\\\ & = & \displaystyle \Pr\left(-0.82 \leq Z \leq 0.47\right) \\\\ \\\\ & = & \displaystyle \Pr\left(Z \leq 0.47\right) - \Pr\left(Z \leq -0.82\right) \\\\ \\\\ & = & 0.681 - 0.2051 \\\\ \\\\ & = & 0.4759 \end{array}$

Therefore, based on the information provided, it is concluded that $$\Pr(11.3 \leq \bar X \leq 12.4) = 0.4759$$.