Probability calculator with mean and standard deviation and sample size

Q: How do you find probability with sample size and mean and standard deviation?

Define your population mean (μ), standard deviation (σ), sample size, and range of possible sample means..

Q: How do you find the probability of a sample mean?

Suppose we draw a sample of size n=16 from this population and want to know how likely we are to see a sample average greater than 22, that is P( > 22)? So the probability that the sample mean will be >22 is the probability that Z is > 1.6 We use the Z table to determine this: P( > 22) = P(Z > 1.6) = 0.0548.

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By Normal we can calculate standard deviation using set of datas (Worksheet for Standard Deviation). We can calculate the Mean and standard deviation using the sample size and probability. using the below formula
Formula for Standard Deviation
sd=√n x p x (1-p)
Formula for Mean
mean= n x p

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Lesson Outcomes
Review of Sampling Distributions
Review of Key Concepts
How do you find probability with sample size and mean and standard deviation?
How do you find the probability of a sample mean?
How do you find the proportion of a population with mean and standard deviation calculator?

Example Problem
A Single dice is throw 450 times and find the standard deviation and mean for the probability of getting 5.

From the above problem
No.of Sample N=450
and probability of getting 5 is 1/6

Now appy N and P in above equation to find Standard Deviation and mean,
SD=√

n x p x (1-p)
SD=√450 x 1/6 x (1-1/6)
SD=7.9057

Mean = n x p
Mean = 450 x 1/6
Mean = 75

Hence the required Standard Deviation and Mean for the given Probability and samples is
SD=7.9057
mean=75

Calculators
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Statistics and probability
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Normal Distribution Calculator

Enter mean, standard deviation and cutoff points and this calculator will find the area under normal distribution curve. The calculator will generate a step by stepexplanation along with the graphic representation of the area you want to find.

EXAMPLES

A normally distributed random variable $X$ has a mean of $20$ and a standard deviation of $4$. Determine the probability that a randomly selected x-value is between $15$ and $22$.

The final exam scores in a statistics class were normally distributed with a mean of $58$ and a standard deviation of $4$. Find the probability that a randomly selected student scored more than $62$ on the exam.

The target inside diameter is $50 \, \text{mm}$ but records show that the diameters follows a normal distribution with mean $50 \, \text{mm}$ and standard deviation $0.05 \, \text{mm}$. An acceptable diameter is one within the range $49.9 \, \text{mm}$ to $50.1 \, \text{mm}$. What proportion of the output is acceptable?

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Lesson Outcomes

By the end of this lesson, you should be able to:

Calculate probabilities using a distribution of sample means.

Review of Sampling Distributions

When is the sample mean normally distributed (or approximately normally distributed)? This happens when either of the two conditions are satisfied:

The parent population is normally distributed, so the sample mean is automatically normally distributed.
The sample size is large, and the Central Limit Theorem implies that the sample mean is normally distributed.

For the mean of draws from a random variable with mean $\mu$ and standard deviation $\sigma$, the following are true:

The mean of the random variable $\bar X$ is $\mu$.
The standard deviation of the random variable $\bar X$ is $\displaystyle{\frac{\sigma}{\sqrt{n}}}$.

Previously, you learned how to use the normal probability applet to convert a $z$-score to a corresponding area under the curve. These skills will be applied again in this activity. The following questions will help you brush up on how to use the applet.

Answer the following questions:

Look at the output from the applet, given above. What is the probability that a standard normal random variable will be below 1.25?

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Look at the output from the applet, given above. What is the probability that a standard normal random variable will be above 1.25?

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Challenge problem: Use the output from the applet above to determine the probability that a standard normal random variable will be between -1.25 and 1.25.

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Use the normal probability applet to find the probability that a standard normal random variable will be greater than -0.75.

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If the mean of a normally distributed random variable is -23 and the standard deviation is 7, what is the probability that the random variable will have a value that is less than -25?

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If the sample mean is normally distributed, then we can consider the sample mean, $\bar x$, as one observation from a normal distribution with mean $\mu$ and standard deviation $\displaystyle{\frac{\sigma}{\sqrt{n}}}$. With this in mind, we can compute the $z$-score for any value of this random variable:

\[z = \frac{\text{value} - \text{mean}}{\text{standard deviation}}=\frac{\bar x-\mu}{\sigma / \sqrt{n} }\]

Notice that we just replaced the “value” with the normal random variable, the “mean” with its mean and “standard deviation” with its standard deviation.

The $z$-score follows a standard normal distribution. That is, it has a mean of 0 and a standard deviation of 1. After we compute the $z$-score, we can use the applet to find the probability that a randomly selected mean will be above or below a given value of $\bar x$.

Worked Example: Finding the Area under a Normal Curve (Based on a Sample Mean)

After finding the $z$-score, we can use the normal probability applet to find the area under the curve (i.e. the probability.) Suppose that a sample of size $n = 4$ has been drawn from a normal population with mean $\mu = 7$ and standard deviation $\sigma = 3$. Since the parent population is normal, we know the sampling distribution of the sample mean $\bar x$ will be normal with mean $\mu = 7$ and standard deviation $\displaystyle{ \frac{\sigma}{\sqrt{n}} = \frac{3}{\sqrt{4}} = 1.5}$. We can use this information to find the probability that $\bar x$ will be greater than 10.

The $z$-score is

\[z=\frac{\bar x-\mu}{\sigma / \sqrt{n} } = \frac{10 - 7}{3 / \sqrt{4} }=2.0\]

The probability that $z$ will be greater than $2.00$ is the area under the standard normal distribution to the right of $2.00$. We find this area using the normal probability applet.

The area to the right of $z = 2.00$ is $0.02275$. This is the probability that the random sample of $n = 4$ items will have a mean that is greater than $10$.

In this example, the sample mean was automatically normally distributed because the parent population was normally distributed. This will always be true, no matter what size sample is drawn.

Worked Example: Finding the Area under a Normal Curve (Based on a Sample Mean)

What do we do if the parent population is not normally distributed? If the sample size is large, then the Central Limit Theorem guarantees that the sample mean will be approximately normally distributed. Based on this, we can still do normal probability calculations for the mean of a random sample.

The distribution of the weekly costs incurred by Global Solutions Unlimited is right skewed. The population mean of the costs is $26,400 and the standard deviation is $23,200. A random sample of $n = 40$ weeks is to be selected, what is the probability that the mean weekly costs will be less than $20,000?

Since the number of observations is large, the Central Limit Theorem assures that the sample mean will follow a normal distribution. The $z$-score for $\bar x =$ $20,000 is:

\[z = \frac{\bar x-\mu}{\sigma / \sqrt{n} } = \frac{20,000-26,400}{23,200 / \sqrt{40}} = -1.745 \]

Using the normal probability applet, we can find the area to the left of this $z$-score:

The area to the left of $z= -1.745$ is $0.04049$. This is the probability that the mean costs for the $n = 40$ weeks will be less than $20,000.

Example: Environmental Clean Up

We will now consider a complete example that shows how these probabilities are used in practice.

The United States Government decided to open some land near a uranium enrichment facility to public use. After a few years of the public hiking, biking, and sometimes even hunting on this land, workers from the facility noticed that there were several unnatural-looking mounds in the earth near the area. Because this land was once used by the facility and nobody knew the origin of these piles, the government closed public access to the land until they could assess if the mounds were safe.

Step 1: Design the study.

Measurements were taken from the mounds to assess one of the contaminants, lead. The tests involved are very expensive. Each sample costs about $600 to process. The Environmental Protection Agency (EPA) has set a “No Action Level” (NAL) for lead. If the mean concentration in the soil of the contaminant is less than the NAL, then the area can be declared safe for public use. If the concentration of the contaminant reaches or exceeds the NAL, the site must be cleaned additionally before it is declared safe. The NAL for lead is 50 milligrams of lead per kilogram of soil (mg/kg).

The hypothesis for this test are:

$H_0:~~\mu = 50 \frac{\text{mg}}{\text{kg}}$

$H_a:~~\mu < 50 \frac{\text{mg}}{\text{kg}}$

In environmental testing, we always assume the site is dirty. That is, our null hypothesis is that the mean level of contamination is at the NAL. We gather data to determine if there is sufficient evidence to support rejecting the null hypothesis.

Step 2: Collect Data

Scientists collected $n = 61$ measurements of the lead concentration in the soil, measured in mg/kg. The data are given in the file Uranium Plant Data-Lead.

Step 3: Describe the Data

Answer the following question:

Step 4: Make Inferences

We assume the null hypothesis is true and we gather evidence against this requirement. We will find the probability that the mean lead concentration is less than $\bar x$ = 29.13 mg/kg, assuming that the true mean lead concentration is $\mu$ = 50 mg/kg. This probability is called the $P$-value.

Since the sample size is large ($n = 61$), we can conclude that the sample mean is normally distributed. So, we can use the normal probability applet to find the probability that the sample mean will be less than 29.13. Historical analyses show that the population standard deviation is $\sigma$ = 24 mg/kg.

First, we compute the $z$-score.

\[z=\frac{\bar x-\mu}{\sigma / \sqrt{n} } = \frac{29.13 - 50}{24 / \sqrt{61} }=-6.79\]

Using the applet, we find the area to the left of $z = -6.79$. This is the $P$-value. It is not clear from the image of the applet, but the area to the left was shaded before the $z$-score was entered.

The $P$-value for this test is $5.6 \times 10^{-12} = 0.000~000~000~005~6$. This is a very small probability. Assuming the null hypothesis is true-that is, the site is unacceptably contaminated-it is very unlikely that we would find such a low mean contamination level among the $n = 61$ randomly selected soil samples.

Since the $P$-value is low, we reject the null hypothesis.

Step 5: Take Action

There is sufficient evidence to suggest that the mean lead level is less than 50 mg/kg. We conclude that the lead concentration in the soil is low enough that it is not a danger to the public. Based on the results of this and other similar test results, the government has reopened public access to this area.

Review of Key Concepts

To compute probabilities involving the sample mean $\bar{x}$ we must first determine if $\bar{x}$ is normally distributed. If it is not, we cannot compute probabilities. There are two cases when $\bar{x}$ is guaranteed to be normally distributed. They are:

The parent population is Normally distributed.
The Central Limit Theorem guarantees the distribution of $\bar{x}$ is Normal if the sample size $n$ is large enough. For this course, we require $n \geq 30$.

The collection of all possible sample means $\bar{x}$ is also a population. The mean of the distribution of sample means is the population mean $\mu$. The standard deviation of the distribution of sample means is $\frac{\sigma}{\sqrt{n}}$, where $\sigma$ is the parent population standard deviation and $n$ is the sample size.
Once we have determined that the sample mean is Normally distributed, we can compute probabilities with $\bar{x}$. We first compute $z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}}$ and then use the Normal Probability Applet to compute probabilities about $z$.

Summary

Remember…

When the distribution of sample means is normally distributed, we can use z-scores and the probability applet to calculate proportions and probabilities. A z-score is calculated as: $\displaystyle{z = \frac{\text{value}-\text{mean}}{\text{standard deviation}} = \frac{\bar x-\mu}{\sigma/\sqrt{n}}}$
The $P$-value is the probability of getting a test statistic at least as extreme as the one you got, assuming $H_0$ is true. A $P$-value is calculated by finding the area under the normal distribution curve that is more extreme (farther away from the mean) than the z-score.

How do you find probability with sample size and mean and standard deviation?

Define your population mean (μ), standard deviation (σ), sample size, and range of possible sample means..

Input those values in the z-score formula zscore = (X̄ - μ)/(σ/√n)..

Considering if your probability is left, right, or two-tailed, use the z-score value to find your probability..

How do you find the probability of a sample mean?

Suppose we draw a sample of size n=16 from this population and want to know how likely we are to see a sample average greater than 22, that is P( > 22)? So the probability that the sample mean will be >22 is the probability that Z is > 1.6 We use the Z table to determine this: P( > 22) = P(Z > 1.6) = 0.0548.

How do you find the proportion of a population with mean and standard deviation calculator?

This is given by the formula Z=(X-m)/s where Z is the z-score, X is the value you are using, m is the population mean and s is the standard deviation of the population. Consult a unit normal table to find the proportion of the area under the normal curve falling to the side of your value.