## AP®︎/College Statistics

• Standard normal table for proportion below
• Standard normal table for proportion above
• Normal distribution: Area above or below a point
• Standard normal table for proportion between values
• Normal distribution: Area between two points
• Finding z-score for a percentile
• Threshold for low percentile
• Normal calculations in reverse

Finding z-score for a percentile

In any normal distribution, we can find the z-score that corresponds to some percentile rank. If we’re given a particular normal distribution with some mean and standard deviation, we can use that z-score to find the actual cutoff for that percentile. In this example, we find what pulse rate represents the top 30% of all pulse rates in a population.

## Want to join the conversation?

• Can someone explain why the z score of 0.53 was used instead of 0.52?
0.52 differed from the ideal value by 0.0015 instead of 0.0019 so it seems like that would have been a better choice. I realize that both give the same answer when rounded to one heartbeat per minute but it still seems odd.(40 votes)
• I don’t agree to the 0.53 either. The question doesn’t state whether she wants at least the top 30% or at max the top 30%, but the former seems reasonable. Choosing 0.53 as the z-value, would mean we ‘only’ test 29.81% of the students.
I would have assumed it would make more sense to choose z=0.52 for that reason, so that we at least cover 30%.(35 votes)
• z = (x – µ)/ σ
So in video’s example,
0.53 = (x – µ)/ σ
0.53 * 11 = x – µ
4.77 + µ = x
• What’s the spreadsheet formula to find the exact z score of a given percentage?
In the video, atwe that the percentage=0.7019 so the z score=0.5 (but we actually want the z score of a 0.7) 3:07(2 votes)
• The exact z score for a given cumulative percentage, in Excel in Office 365, is either
=NORMSINV(percentage)
or
=NORM.S.INV(percentage)
So the exact z score for a cumulative percentage of 0.7 is either
=NORMSINV(0.7)
or
=NORM.S.INV(0.7)
Have a blessed, wonderful day!(11 votes)
• How did you know it had to be at least 70% ?(5 votes)
• Where do you find the z-score table(4 votes)
• http://www.z-table.com/
You are able to find z-score table on this website :)(3 votes)
• No table interpolation any more? That’s surprising to me. I understand that it’s irrelevant on this example, with a rounding to the nearest natural number; but the concept is really important, and also, I believe, valuable in a pre-calculus context.(3 votes)
• How do you find the mean and standard deviation given a percentile and a value? For example, if you are given that 16 is the 25th percentile and 40 and the 97.5% for a normal distribution.(3 votes)
• Where can I get a down load of the z-scores?(2 votes)
• I know its maybe too much, but wouldn’t more correct answer be in z table (0.6985+0.7019)/2 = 0.7002, which would be exactly 0,7002 and that gives us z score of 5,25(1 vote)
• Averaging the two scores would give you a more accurate z-score, but it’s important to note that averaging the z-scores does not average the percentiles, so it wouldn’t be exactly 0.7002. It’s a good estimate in this case because the scores are so close together, and the actual value with a z score of .525 is marginally different. The other thing to note is that we’re rounding to the nearest whole number pulse rate, so a z-score that’s 0.0019 off is unlikely to affect that answer.(3 votes)
• Can someone explain how to get .52 if you aren’t given a table?
I am taking a teaching test soon and we won’t be given a table.
Thanks(1 vote)
• Are you sure? There is no way to do it by hand. The only ways are to use the table or use a calculator.(2 votes)

## Video transcript

– [Instructor] The distribution of resting pulse rates of all students at Santa Maria High School was approximately normal with mean of 80 beats per minute and standard deviation of nine beats per minute. The school nurse plans to provide additional screening to students whose resting pulse rates are in the top 30% of the students who were tested. What is the minimum resting pulse rate at that school for students who will receive additional screening? Round to the nearest whole number. If you feel like you know how to tackle this, I encourage you to pause this video and try to work it out. All right, now let’s work this out together. They’re telling us that the distribution of resting pulse rates are approximately normal. So we could use a normal distribution. And they tell us several things about this normal distribution. They tell us that the mean is 80 beats per minute. So that is the mean right over there. And they tell us that the standard deviation is nine beats per minute. So on this normal distribution, we have one standard deviation above the mean, two standard deviations above the mean, so this distance right over here is nine. So this would be 89. This one right over here would be 98. And you could also go standard deviations below the mean, this right over here would be 71, this would be 62, but what we’re concerned about is the top 30% because that is who is going to be tested. So there’s gonna be some value here, some threshold. Let’s say it is right over here, that if you are at that score, you have reached the minimum threshold to get an additional screening. You are in the top 30%. So that means that this area right over here is going to be 30% or 0.3. So what we can do, we can use a z-table to say for what z-score is 70% of the distribution less than that. And then we can take that z-score and use the mean and the standard deviation to come up with an actual value. In previous examples, we started with the z-score and were looking for the percentage. This time we’re looking for the percentage. We want it to be at least 70% and then come up with the corresponding z-score. So let’s see, immediately when we look at this, and we are to the right of the mean, and so we’re gonna have a positive z-score. So we’re starting at 50% here. We definitely want to get to the 67%, 68, 69, we’re getting close and on our table this is the lowest z-score that gets us across that 70% threshold. It’s at 0.7019. So it definitely crosses the threshold. And so that is a z-score of 0.53. 0.52 is too little. So we need a z-score of 0.53. Let’s write that down. 0.53, right over there, and we just now have to figure out what value gives us a z-score of 0.53. Well, this just means 0.53 standard deviations above the mean. So to get the value, we would take our mean and we would add 0.53 standard deviation. So 0.53 times nine. And this will get us 0.53 times nine is equal to 4.77 plus 80 is equal to 84.77. 84.77 and they want us to round to the nearest whole number. So we will just round to 85 beats per minute. So that’s the threshold. If you have that resting heartbeat, then the school nurse is going to give you some additional screening. You are in the top 30% of students who are tested.

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