Sampling Distribution of Sample Proportion: Videos & Practice Problems

In statistics, the binomial probability model approximates normal distribution when both $np5$ and $n(1p)5$ hold. The sample proportion $p̂$ is derived from successes $x$ divided by trials $n$. The mean of $p̂$ equals $p$, while the standard deviation is $\sqrt{\frac{p}{(}}/n$. Understanding these concepts is crucial for constructing confidence intervals and conducting hypothesis tests.

Using the Normal Distribution to Approximate Binomial Probabilities

Earlier in this course, we learned how to find probabilities using the binomial formula. If we had n trials with two possible outcomes resulting in x successes and a probability of success p, we could plug all of that information into our binomial formula to find the probability that we were looking for. But as n, our number of trials, or our sample size, gets larger and larger, using the binomial formula can become a much more tedious process. But luckily, there's an easier way. If we were to do a bunch of binomial experiments and plot all of our successes from each sample, we would get a graph that looks like this.

And we can see that this looks just like our normal distribution. And as it turns out, if both np and nq are greater than or equal to five, we can use our normal distribution in order to approximate a binomial probability. So we can use a z score to find a probability instead of having to use the binomial formula. Now this is the process that we're going to walk through together here. So let's go ahead and get started by jumping right into our example.

In this problem, we're told that the probability of someone voting for a particular candidate in a two-person election is 56%, and we want to find here the probability that more than 60 of a sample of 100 people vote for that particular candidate. Now, if we were to do this using the binomial formula, since we're trying to find the probability that more than 60 out of the sample of 100 people voted for this candidate, we would have to find the probability that x is equal to sixty, sixty-one, sixty-two, sixty-three, all the way up to 100 using the binomial formula 40 different times. And this definitely doesn't sound like a process that we want to do. But remember that if np and nq are greater than or equal to five, we don't have to do that, and we can instead use the normal distribution. So let's check if that's true for this particular problem, that np and nq are greater than or equal to five.

Now here n is equal to 100, that's our total sample size, and we have a p value of 56%. So taking n times p, that's 100 times 0.56. That is definitely greater than or equal to five. That gives us 56. Then if we take n times q that's 100 times 0.44 since that's just one minus p that gives us 44 which is also greater than or equal to five.

Now since both of those are true we know that we can use our normal distribution here in order to approximate this probability. So in order to do that, we need to find our z score. Remember from working with our binomial variable in the past that it has a mean of n times p and a standard deviation that's equal to the square root of npq. So that means that our z score is going to be x minus that mean, that np, over the standard deviation, that's the square root of npq. So to find our z score, we just need to plug all of this information in and do a bit of algebra.

Now, looking at this equation, we obviously need to know what x is. And since in our problem, we're trying to find the probability that more than 60 people voted for this candidate, you may be assuming here that 60 should get plugged in for x. But since we are using our normal distribution to approximate our binomial distribution, we have to make what's called a continuity correction because we're using our continuous normal distribution to approximate our discrete binomial distribution. So looking at our graph here, 60 represents the midpoint of this interval that has a total width of one. That means that either endpoint of this interval can be found by subtracting and adding 0.5.

So that lower endpoint is 59.5, and that upper endpoint is 60.5. In order to make this continuity correction and account for this entire interval, we're going to have to plug in 59.5 or 60.5 as x instead of 60. But how do we know which one to choose? Well, it's based on the probability that we're trying to find. And here I have this cheat sheet that helps us determine whether we need to add or subtract point zero five based on what's told to us in the problem and the probability that we're trying to find.

In this particular problem, we're trying to find the probability that more than 60 people are doing something. So looking at this bottom row down here, since we're looking at the probability that x is greater than 60, we need to go ahead and do our continuity correction by adding point five. So looking at our graph here, this makes sense because accounting for that whole interval, we want to find the probability that x is greater than that entire interval. So using that upper endpoint, we would want to add 0.5 to x. So we're going to use 60.5 when we actually calculate our z score.

So now that we know that we needed to add 0.5 to do our continuity correction here, we can go ahead and actually find z. So coming back up to our problem here, z = x - n × p n × p × q So using that corrected value, 60.5 minus n, that's 100, times p, which is that 0.56, dividing that by the square root of npq, that's 100 times 0.56 times 0.44. Now actually working out this algebra ends up giving us a z score of 0.907. So we know that based on this z score we can now find our probability going to this next step here since we're looking for the probability that x is greater than 60.5.

Remember, having corrected for continuity. This then corresponds to the probability that z is greater than 0.907 and from this point, we know that we can either use a calculator or a z table to find this probability. When you do that, you should end up getting that this probability is equal to 0.182. So this is the probability that more than 60 people out of the sample of 100 vote for that one particular candidate. So here, we've successfully used our normal distribution to approximate our binomial probability, instead of having to use the binomial formula 40 different times.

So we're going to continue getting practice with this coming up in the next couple of videos. Let us know if you have any questions, and I'll see you there.

A bank analyzes customer adoption of its new mobile banking app. Historically, 45% of customers use online banking services. Use a normal distribution to approximate the probability that between 62 and 70 customers out of a sample of 100 will adopt the online banking service.

• A. 0.00165
• B. 0.000463
• C. 0.99
• D. 0.01

Answer: (B)

A previous study found that $80\%$ of people preferred drinking Pepsi over Coca Cola. Use a normal distribution to approximate the probability that a random sample of $100$ people reveals $60$ people or more preferring Pepsi. 

• A. 0.8
• B. 0.6
• C. 0
• D. 1

Answer: (D)

A previous study found that $80\%$ of people preferred drinking Pepsi over Coca Cola. Use a normal distribution to approximate the probability that, from this same random sample of $100$ people, that between $10$ and $11$ people prefer Coca Cola.

• A. 0.0125
• B. 0.9875
• C. 0.8
• D. 0.105

Answer: (A)

Using the Normal Distribution to Approximate Binomial Probabilities Example 1

If you haven't already, go ahead and give this problem a try on your own before checking back in with me. Here we're told that in 2023, about 35,000,000 cars were recalled out of approximately 94,000,000 cars produced, so that's 35,000,000 out of that 94,000,000. We're told here that a used car dealer has 76 cars that were produced in 2023. We're asked to use a normal distribution to approximate the probability that more than half of these cars were recalled. In this problem, we are asked to use a normal distribution to approximate a binomial probability.

Even though we're told explicitly to do that, we want to make sure that we can do so by checking that NP and NQ are both greater than or equal to five. Now in this case, since we are not explicitly given p, we have to calculate it by taking that 35,000,000 and dividing it by 94,000,000. Because they're both in millions, I can just take 35 and divide it by 94 to give me a p-value of 0.372, which then means that q is equal to one minus 0.372, that's 0.628. Now if I take p and multiply it by n, that is 76 for this particular problem.

That is greater than or equal to five as is n times q. So because both of these things are true, we know that we can use our normal distribution to approximate this binomial probability. Here we're trying to find the probability that more than half of those 76 cars were recalled. Now, half of 76, if we take 76 and divide it by two, that is 38. So you want to find the probability that x is greater than 38.

Remember that we have to do a continuity correction here because we are using a continuous normal distribution to approximate our discrete binomial distribution. So we need to go ahead and add 0.5 in order to correct here. So we're really looking at the probability that X is greater than 38.5. Now in order to find this probability, we need to find our z score. Remember that z is going to be equal to x−n·pn·p·q.

So plugging in all of our values here, we're plugging in 38.5 for x because we are using that corrected value minus np. So that is 76 times the calculated p value of 0.372 divided by the square root of (76 times p 0.372 times 0.628). Working this out algebraically will give us a z value here of 2.427. So then this probability that x is greater than 38.5, using that z score, is equal to the probability that z is greater than 2.427. Now we can either use a z table or a calculator from this point to find this probability and we should end up finding that this is equal to 0.00761.

This 0.00761 is the probability that more than half of this particular dealer's 76 cars were recalled. Feel free to let us know if you have any questions and let's keep going.

Sampling Distribution of Sample Proportion

Earlier in this chapter, we learned that the binomial variable x has an approximately normal distribution when both np and nq are greater than or equal to five. Now, for a binomial experiment, if instead of just looking at the number of successes, if we took the number of successes x and divided that by the total number of trials n, we would get a proportion of our successes referred to as the sample proportion denoted by the variable p̂. Now taking a bunch of random samples here gives us the sample distribution of p̂, which has a lot of similarities to our binomial distribution, including being approximately normal when np and nq are greater than or equal to five. Now, here we're going to take a look at why that is and learn some additional information about our sample proportion, p̂. So let's go ahead and get started here by jumping right into our example.

Here we're told that you and several classmates take turns flipping a coin 10 times each. Now the distribution of heads, which we're referring to as our successes here, is shown in the graph below. And we're asked to find the mean and standard deviation of the sampling distribution for p̂. So let's dive into everything going on in this problem. You and several classmates are all flipping coins, so this is a bunch of random samples.

Now since you're each doing this 10 times, that gives us our total number of trials n. And the distribution of heads, those are our successes, x, are shown in this graph below. This is our binomial distribution. Now because we want to find the mean and standard deviation of the sampling distribution for p̂, we need to understand what p̂ is. Earlier I mentioned that p̂ is the actual number of successes out of the total number of trials.

So that means that p̂ is equal to x, those successes, divided by the number of trials n. Now, coming back over here to our left side with our binomial distribution, we can see our successes plotted here on the x axis. Now looking at all of these values for x, if we took all of them and divided by n, our total number of trials, that would give us our sample proportion, p̂. Now doing that for all of these gives us the distribution of our sample proportions. So taking three and dividing that by 10 will give me 0.3.

Then doing that for all of the successive values is going to give me 0.4, 0.5, 0.6, and 0.7. Now we can see that the shape of this graph is identical to that of our binomial distribution because it's the exact same but with a rescaled x axis, having divided by that total number of trials, in this case, 10. So because our binomial distribution is approximately normal, so is the distribution of our sample proportion. Now because here we were able to just divide all of these values by n, we can take the same approach when finding the mean and standard deviation of p̂.

We can take the mean and standard deviation of our binomial variable and just divide them by n, the total number of trials. So for our mean here, if we take np, the mean of our binomial variable x, and divide that by n, that gives us that the mean of p̂ is just equal to p. Then doing the same thing for our standard deviation sigma, taking the square root of npq dividing that by n gives us pqn. Now we can go ahead and calculate those values here and remember that we're still using the same values of n and p. This is still the exact same scenario that we are working with for our binomial distribution.

So we still have an n value of 10 and p is 0.5. That's the probability of getting heads. Now just getting out of the way here, if we go ahead and look at mu of p̂, remember that that's just p. So since p is 0.5, the mean of p̂ is also 0.5. Then calculating our standard deviation here, sigma of p̂, that's equal to pqn.

So that's 0.5 times q, which is also 0.5, dividing that by n, our total number of trials, in this case 10. Working this out ends up giving us a value here of 0.158. Now looking at each of these values, I can see that they are just the same as those for our binomial variable, but just divided by 10. So what are important takeaways here? Well, our sample proportion p̂ is equal to x over n.

To get our sample distribution of p̂, we just took everything going on with our binomial distribution and divided it by our total number of trials n. Because the distribution of p̂ is basically just our binomial distribution with a rescaled x axis, that means that it can be assumed to be approximately normal based on the same criteria, np and nq being greater than or equal to five. And we're going to continue using this information as we continue on in statistics and specifically when we start to construct confidence intervals for proportions. Now feel free to let us know if you have any questions here, and I'll see you in the next one.

A market research firm is studying customer satisfaction for a food delivery service. Based on past data, $85\%$ of customers rate the service as "satisfactory". The firm randomly surveys groups of 250 customers. Find the mean ${\mu }_{\hat{p}}$ and standard deviation ${\sigma }_{\hat{p}}$ of the sampling distribution for $\hat{p}$. What would the shape of the sampling distribution be?

• A. ${\mu }_{\hat{p}}=0.34$${\sigma }_{\hat{p}}=0.0300$The sampling distribution is normal
• B. $\mu_{p̂}=0.85$$\sigma_{p̂}=0.0226$The sampling distribution is skewed
• C. $\mu_{p̂}=0.85$$\sigma_{p̂}=0.0226$The sampling distribution is normal
• D. $\mu_{p̂}=0.34$$\sigma_{p̂}=0.0300$The sampling distribution is skewed

Answer: (C)

Here’s what students ask on this topic:

How can the normal distribution be used to approximate a binomial distribution in sampling?

In sampling, the normal distribution can approximate a binomial distribution when both $n_p$ and $n_q$ are greater than or equal to five. This condition ensures that the binomial distribution is sufficiently symmetric and can be approximated by the bell-shaped curve of the normal distribution. The mean of the binomial distribution is $n_p$, and the standard deviation is $\sqrt{n_p{q}_{}}$. Using these values, a z-score can be calculated to find probabilities using the normal distribution, simplifying the process compared to using the binomial formula multiple times.

What is the sample proportion and how is it calculated?

The sample proportion, denoted as $\hat{p}$, is the ratio of the number of successes to the total number of trials in a sample. It is calculated using the formula $\frac{x}{n}$, where $x$ is the number of successes and $n$ is the total number of trials. The sample proportion provides a way to express the outcome of a binomial experiment as a percentage, making it easier to compare results across different sample sizes. Understanding the sample proportion is essential for constructing confidence intervals and performing hypothesis testing in statistics.

Why is the continuity correction necessary when using the normal distribution to approximate a binomial distribution?

The continuity correction is necessary when using the normal distribution to approximate a binomial distribution because the normal distribution is continuous, while the binomial distribution is discrete. This correction accounts for the difference between the two types of distributions. When calculating probabilities, the continuity correction involves adjusting the discrete value by adding or subtracting 0.5. This adjustment ensures that the approximation is more accurate by considering the entire interval around the discrete value. For example, if finding the probability that $x$ is greater than a certain value, 0.5 is added to the value to account for the interval.

How do you find the mean and standard deviation of the sampling distribution for the sample proportion?

To find the mean and standard deviation of the sampling distribution for the sample proportion $\hat{p}$, use the following formulas: The mean of the sample proportion is equal to the population proportion $p$. The standard deviation is calculated using the formula $\sqrt{\frac{\mathrm{pq}}{n}}$, where $q$ is $(1-p)$ and $n$ is the sample size. These calculations are crucial for understanding the variability and central tendency of the sample proportion, which are important for statistical inference.