Sampling Distribution of Sample Proportion: Videos & Practice Problems

In binomial experiments, when both n⁢p⁢5 and n(1-p)⁢5 hold, the binomial distribution can be approximated by a normal distribution. The sample proportion p^ is defined as x/n, where x is the number of successes. The mean of p^ is equal to p, and the standard deviation is p(1-p)/n. This understanding is crucial for constructing confidence intervals for proportions.

Using the Normal Distribution to Approximate Binomial Probabilities

In probability theory, the binomial formula is a useful tool for calculating the likelihood of a certain number of successes in a fixed number of trials, where each trial has two possible outcomes. However, as the number of trials (n) increases, using the binomial formula can become cumbersome. Fortunately, when certain conditions are met, we can simplify this process by using the normal distribution to approximate binomial probabilities.

To apply this approximation, we need to ensure that both np and nq are greater than or equal to 5, where p is the probability of success and q is the probability of failure (calculated as q = 1 - p). For example, if we have a scenario where the probability of voting for a candidate is 56% (or 0.56), and we are sampling 100 people, we can calculate:

np = 100 * 0.56 = 56 and nq = 100 * 0.44 = 44. Since both values are greater than 5, we can proceed with the normal approximation.

The next step involves calculating the z-score, which is a measure of how many standard deviations an element is from the mean. The mean of our binomial distribution is given by μ = np, and the standard deviation is σ = \sqrt{npq}. The z-score can be calculated using the formula:

z = \frac{x - μ}{σ}

In our example, we want to find the probability that more than 60 people vote for the candidate. To do this, we need to apply a continuity correction because we are using a continuous normal distribution to approximate a discrete binomial distribution. This involves adjusting our value of x by adding or subtracting 0.5. Since we are looking for the probability that x is greater than 60, we add 0.5, resulting in x = 60.5.

Now we can calculate the z-score:

z = \frac{60.5 - (100 * 0.56)}{\sqrt{100 * 0.56 * 0.44}} = \frac{60.5 - 56}{\sqrt{24.64}} = \frac{4.5}{4.964} \approx 0.907

With the z-score calculated, we can now find the corresponding probability using a z-table or calculator. The probability that z is greater than 0.907 is approximately 0.182. Therefore, the probability that more than 60 out of 100 people vote for the candidate is about 18.2%.

This method of using the normal distribution to approximate binomial probabilities is particularly useful for large sample sizes, allowing for quicker calculations without the need to evaluate the binomial formula multiple times.

The probability of someone voting for a particular candidate in a two-person election is $45\%$ . Use a normal distribution to approximate the probability that between $62$ and $70$ people out of a sample of $100$ vote for the candidate. 

• 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

In 2023, approximately 35,000,000 cars were recalled from a total of about 94,000,000 cars produced. A used car dealer has 76 cars produced in that year, and we want to determine the probability that more than half of these cars were recalled using a normal distribution to approximate a binomial probability.

To begin, we need to calculate the probability of a car being recalled, denoted as p. This is done by dividing the number of recalled cars by the total number of cars produced:

p = \frac{35,000,000}{94,000,000} = \frac{35}{94} \approx 0.372

Consequently, the probability of a car not being recalled, q, is:

q = 1 - p = 1 - 0.372 = 0.628

Next, we check the conditions for using the normal approximation by calculating np and nq:

np = 76 \times 0.372 \approx 28.272 \quad \text{and} \quad nq = 76 \times 0.628 \approx 47.728

Both values are greater than 5, confirming that we can use the normal distribution for this approximation.

We are interested in finding the probability that more than half of the 76 cars were recalled. Half of 76 is 38, so we need to find the probability that X is greater than 38. To account for the continuity correction, we adjust this to finding the probability that X is greater than 38.5.

To find this probability, we calculate the z-score using the formula:

z = \frac{x - np}{\sqrt{npq}}

Substituting the values:

z = \frac{38.5 - (76 \times 0.372)}{\sqrt{76 \times 0.372 \times 0.628}} = \frac{38.5 - 28.272}{\sqrt{76 \times 0.372 \times 0.628}} \approx 2.427

Now, we need to find the probability that Z is greater than 2.427. This can be done using a z-table or a calculator. The resulting probability is:

P(Z > 2.427) \approx 0.00761

This means there is approximately a 0.761% chance that more than half of the dealer's 76 cars were recalled. This low probability indicates that it is quite unlikely for more than half of the cars to have been recalled.

Sampling Distribution of Sample Proportion

In the context of binomial experiments, the sample proportion, denoted as \( \hat{p} \), is derived by dividing the number of successes \( x \) by the total number of trials \( n \). This proportion provides valuable insights into the distribution of successes across multiple trials. When both \( np \) and \( nq \) (where \( q = 1 - p \)) are greater than or equal to 5, the distribution of \( \hat{p} \) can be approximated as normal, similar to the binomial distribution.

To illustrate this, consider a scenario where several students flip a coin 10 times each. The number of heads obtained represents the successes in this experiment. The distribution of these successes can be visualized, and by calculating the sample proportion \( \hat{p} \) for each student, we can observe the resulting distribution of \( \hat{p} \). For instance, if one student gets 3 heads, their sample proportion would be \( \hat{p} = \frac{3}{10} = 0.3 \). Continuing this for all students yields a range of sample proportions, which collectively form a distribution that mirrors the shape of the original binomial distribution, albeit rescaled along the x-axis.

To find the mean and standard deviation of the sampling distribution of \( \hat{p} \), we can utilize the properties of the binomial distribution. The mean of \( \hat{p} \) is simply equal to \( p \), the probability of success. Therefore, if the probability of getting heads (success) is \( p = 0.5 \), the mean of \( \hat{p} \) is also \( 0.5 \). The standard deviation of \( \hat{p} \) can be calculated using the formula:

\(\sigma_{\hat{p}} = \sqrt{\frac{pq}{n}} = \sqrt{\frac{0.5 \times 0.5}{10}} = 0.158\).

This calculation shows that the standard deviation of the sample proportion is influenced by both the probability of success and the number of trials. The key takeaway is that the distribution of the sample proportion \( \hat{p} \) retains the characteristics of the binomial distribution, allowing us to apply normal approximation methods when constructing confidence intervals for proportions in future statistical analyses.

Here’s what students ask on this topic:

What is the sampling distribution of the sample proportion?

The sampling distribution of the sample proportion, denoted as p̂, is the distribution of sample proportions obtained from all possible samples of a given size from a population. It is approximately normal if the conditions np ≥ 5 and n(1-p) ≥ 5 are met. The mean of the sampling distribution of p̂ is equal to the population proportion p, and the standard deviation is given by the formula:

p(1-p)n

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

To find the mean and standard deviation of the sampling distribution of the sample proportion p̂, use the following formulas:

Mean (μ_p̂):

μ=p

Standard Deviation (σ_p̂):

p(1-p)n

Here, p is the population proportion, and n is the sample size. These formulas assume that the conditions np ≥ 5 and n(1-p) ≥ 5 are satisfied.

When can the normal distribution be used to approximate the binomial distribution?

The normal distribution can be used to approximate the binomial distribution when the sample size n is large enough such that both np and n(1-p) are greater than or equal to 5. This is known as the rule of thumb for normal approximation. When these conditions are met, the binomial distribution is approximately normal, and we can use the z-score to find probabilities instead of the binomial formula.

What is the continuity correction in the context of normal approximation to the binomial distribution?

The continuity correction is an adjustment made when using the normal distribution to approximate a binomial distribution. Since the binomial distribution is discrete and the normal distribution is continuous, we adjust by adding or subtracting 0.5 to the discrete x value. For example, to find the probability that x is greater than 60, we use 60.5 in the z-score formula. This correction helps improve the accuracy of the approximation.

How do you calculate the z-score for a sample proportion?

To calculate the z-score for a sample proportion p̂, use the following formula:

z=p̂-pp(1-p)n

Here, p̂ is the sample proportion, p is the population proportion, and n is the sample size. This z-score can then be used to find probabilities using the standard normal distribution.