why is the formula for Standard deviation either sqrt(np(1-p)) OR its sqrt(p(1-p)/n)?

It depends on what quantity you’re taking the standard deviation of. In a binomial distribution, the first formula you wrote is the standard deviation of the *number* of successes, while the second formula you wrote is the standard deviation of the sample *proportion* of successes. Have a blessed, wonderful day!

how do you tell if the sampling distribution is describing proportions or means

Proportions would sound like "40% of the population knows morse code," where it's a "yes or no" situation. They either do have a certain trait/item or don't. In the case of proportions, the p given is the mean, as seen in the problem on this page. For means, it would be more like "The average age of people that know morse code is 50 years old," where there's a range of possible values. I hope this helps! (The situations I made up don't contain real data lol)

In a set of 10,000 invoices,it is known that 500 contain errors.If 100 of the 10,000 invoices are randomly selected,what is the probability that the sample proportion of invoices with errors will exceed 0.08?

First, calculate your population proportion. p = 500/10,000 = 0.05 Your sample size is 100. Next, check for normality. np >= 10 AND n(1-p) >= 10 100*0.05 = 5 which is NOT >= 10. 100*0.95 = 95 which IS >= 10. The sample distribution of sample proportions violates normality.

Main content

Course: AP®︎/College Statistics > Unit 9

Lesson 4: Sampling distributions for sample proportions

Sampling distribution of a sample proportion example

Google Classroom

Here's the type of problem you might see on the AP Statistics exam where you have to use the sampling distribution of a sample proportion.

Example: Proportions in polling results

According to the US Census Bureau's American Community Survey,

87 %

of Americans over the age of 25 have earned a high school diploma. Suppose we are going to take a random sample of

200

Americans in this age group and calculate what proportion of the sample has a high school diploma.

What is the probability that the proportion of people in the sample with a high school diploma is less than $85 %$ ‍?

Let's solve this problem by breaking it down into smaller parts.

Part 1: Establish normality

Note: The sampling distribution of a sample proportion

\hat{p}

is approximately normal as long as the expected number of successes and failures are both at least

10

Question A (Part 1)

What is the expected number of people in the sample with a high school diploma?

people

Question B (Part 1)

What is the expected number of people in the sample without a high school diploma?

people

Question C (Part 1)

Is the sampling distribution of $\hat{p}$ ‍ approximately normal?

Part 2: Find the mean and standard deviation of the sampling distribution

The sampling distribution of a sample proportion

\hat{p}

has:

\begin{aligned} μ_{\hat{p}} & = p \\ σ_{\hat{p}} & = \sqrt{\frac{p (1 - p)}{n}} \end{aligned}

Note: For this standard deviation formula to be accurate, our sample size needs to be

10 %

or less of the population so we can assume independence.

Question A (Part 2)

What is the mean of the sampling distribution of $\hat{p}$ ‍?

μ_{\hat{p}} =

Question B (Part 2)

What is the standard deviation of the sampling distribution of $\hat{p}$ ‍?
You may round your answer to three decimal places.

σ_{\hat{p}} =

Part 3: Use normal calculations to find the probability in question

What is the probability that the proportion of people in the sample with a high school diploma is less than $85 %$ ‍?

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Ollie
Posted 3 years ago. Direct link to Ollie's post “why is the formula for St...”
why is the formula for Standard deviation either sqrt(np(1-p)) OR its sqrt(p(1-p)/n)?
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(3 votes)
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- Ian Pulizzotto
  Posted 2 years ago. Direct link to Ian Pulizzotto's post “It depends on what quanti...”
  It depends on what quantity you’re taking the standard deviation of. In a binomial distribution, the first formula you wrote is the standard deviation of the number of successes, while the second formula you wrote is the standard deviation of the sample proportion of successes.
  
  Have a blessed, wonderful day!
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  (19 votes)
dennisj
Posted 4 years ago. Direct link to dennisj's post “how do you tell if the sa...”
how do you tell if the sampling distribution is describing proportions or means
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(2 votes)
Answer
- Brad Barakat
  Posted 4 years ago. Direct link to Brad Barakat's post “Proportions would sound l...”
  Proportions would sound like "40% of the population knows morse code," where it's a "yes or no" situation. They either do have a certain trait/item or don't. In the case of proportions, the p given is the mean, as seen in the problem on this page.
  For means, it would be more like "The average age of people that know morse code is 50 years old," where there's a range of possible values.
  I hope this helps!
  (The situations I made up don't contain real data lol)
  Button navigates to signup page
  (3 votes)
rdeyke
Posted 5 years ago. Direct link to rdeyke's post “Sorry, but using a normal...”
Sorry, but using a normal distribution to solve this problem gives incorrect results.

The sampling distribution is a binomial distribution. Using the formula for binomial distributions, one can determine that exactly 85% of the sample has a high school diploma is a whopping 0.0561. It therefore makes a huge difference if we are looking at the probability that the 85% or less of the sample have a high school diploma, or if we are looking at the probability that strictly less than 85% have a diploma. Using the binomial distribution formula again, the former gives a value of 0.2273 and the latter 0.1711. Since the question is asking about P(p^<0.85), 0.17 is in fact the correct answer.
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(3 votes)
Answer
- daniella
  Posted a month ago. Direct link to daniella's post “The sampling distribution...”
  The sampling distribution of a sample proportion is based on the binomial distribution. The binomial distribution provides the exact probabilities for the number of successes in a fixed number of independent Bernoulli trials (like success/failure or yes/no).
  
  When the sample size is large, the sampling distribution of the sample proportion can be approximated by a normal distribution due to the Central Limit Theorem. However, when dealing with exact probabilities, especially at specific proportions like "exactly 85%", it's crucial to use the binomial distribution rather than the normal approximation.
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  (1 vote)
jplatt
Posted 4 years ago. Direct link to jplatt's post “What is the best way to f...”
What is the best way to find standard deviarion.
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(0 votes)
Answer
- daniella
  Posted a month ago. Direct link to daniella's post “For the sampling distribu...”
  For the sampling distribution of a sample proportion, the standard deviation (SD) can be calculated using the formula:
  
  SD = sqrt(p(1 - p) / n)
  
  where p is the population proportion and n is the sample size.
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  (1 vote)
Rorisang Lesaoana
Posted 6 years ago. Direct link to Rorisang Lesaoana's post “In a set of 10,000 invoic...”
In a set of 10,000 invoices,it is known that 500 contain errors.If 100 of the 10,000 invoices are randomly selected,what is the probability that the sample proportion of invoices with errors will exceed 0.08?
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(0 votes)
Answer
- Cory Cascalheira
  Posted 5 years ago. Direct link to Cory Cascalheira's post “First, calculate your pop...”
  First, calculate your population proportion.
  p = 500/10,000 = 0.05
  
  Your sample size is 100.
  
  Next, check for normality.
  np >= 10 AND n(1-p) >= 10
  100*0.05 = 5 which is NOT >= 10.
  100*0.95 = 95 which IS >= 10.
  
  The sample distribution of sample proportions violates normality.
  Comment on Cory Cascalheira's post “First, calculate your pop...”
  (4 votes)
Mohamed Ibrahim
Posted 4 years ago. Direct link to Mohamed Ibrahim's post “Why do we need to prove i...”
Why do we need to prove independence to get the sample proportion standard deviation and not to get the mean ?
Button navigates to signup pageComment on Mohamed Ibrahim's post “Why do we need to prove i...”
(1 vote)
Answer
- daniella
  Posted a month ago. Direct link to daniella's post “Independence is a crucial...”
  Independence is a crucial assumption for using the standard deviation formula of the sample proportion. This assumption ensures that the sampling distribution behaves similarly to the binomial distribution.
  
  The mean does not require the same independence assumption because the expected value of the sample proportion is directly related to the population proportion and the sample size, and it doesn't rely on individual outcomes being independent.
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  (1 vote)
ariannab
Posted 3 years ago. Direct link to ariannab's post “I don't really know what ...”
I don't really know what I'm doing... How do I find the answer to something using only the mean, the standard deviation, and the total population while knowing there are only two possible outcomes in the total population.
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(1 vote)
Answer
- daniella
  Posted a month ago. Direct link to daniella's post “You'd use the properties ...”
  You'd use the properties of the normal distribution. The mean (μ) represents the center of the distribution, and the standard deviation (σ) represents the spread.
  
  If there are only two possible outcomes and you want to find the probability of one of them, you'd use the Normal CDF function in Excel or similar tools.
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  (1 vote)
Bryan
Posted 4 years ago. Direct link to Bryan's post “I'm still confused as to ...”
I'm still confused as to how we can use normal calculations, like a z-table.

The sampling distribution (of sample proportions) is a discrete distribution, and on a graph, the tops of the rectangles represent the probability.
The z-table/normal calculations gives us information on the area underneath the normal curve, since normal dists are continuous.

So we have a sampling dist, and we want to find the probability that we get a sample proportion that is less than 0.7. We know that the dist is approximately normal, and we have it's mean, and SD.
The probability that sample proportion < 0.7 is the tops of all the rectangles below 0.7 summed up for the sampling distribution.
But, for the normal dist (density curve) that approximates our sampling dist, using normalcdf on a calculator or a z-table gives us the proportion of the area under the curve that is < 0.7.

So how is the tops of all the rects below 0.7 summed up equal to the area of the rectangles (area under the normal curve) that is below 0.7?
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(1 vote)
Answer
- Victor Gutierrez
  Posted 4 years ago. Direct link to Victor Gutierrez's post “"The sampling distributio...”
  "The sampling distribution (of sample proportions) is a discrete distribution, and on a graph, the tops of the rectangles represent the probability.
  The z-table/normal calculations gives us information on the area underneath the normal curve, since normal dists are continuous."
  
  But we have not here barrs whre the top of each bar represent a probability. What we have here is a dot plot, and the height of each "bar" in a dot plot doesn´t represent a probability.
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  (0 votes)
Deepti Bist
Posted 4 years ago. Direct link to Deepti Bist's post “Hi, I do not have a calcu...”
Hi, I do not have a calcultor as used in this exercise. Can someone explain, how can I use Excel to get the Normal CDF. I have tried and my answer is not the same. Many Thanks
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Answer
- daniella
  Posted a month ago. Direct link to daniella's post “To calculate the Normal C...”
  To calculate the Normal CDF in Excel, you can use the NORM.DIST function.
  
  For the cumulative probability to the left of a specific value (e.g., less than 0.85), use:
  =NORM.DIST(0.85, mean, standard deviation, TRUE)
  
  For the cumulative probability to the right of a specific value, use: =1 - NORM.DIST(0.85, mean, standard deviation, TRUE)
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  (1 vote)
isaac.uzzle
Posted 3 years ago. Direct link to isaac.uzzle's post “Question 2 Part A has a t...”
Question 2 Part A has a typo, the question on my screen clearly said round to 3 decimal places, but when I entered the correct answer it told me it was incorrect. So I checked the explanation and it had the answer rounded to the 4th decimal place instead of the third. I put the 4th decimal place answer in and it worked. You need to fix that because its going to confuse people who follow directions but then get the problem wrong.
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(0 votes)
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