What do you do if your z-score is negative ?

Then your value is below the mean. Because the table does not have negative values, so assuming that Z follows a normal distribution, you can use P(Z<-x) = 1-P(Z<x)

In one of the practice problems, the z-score was 2. This means that it's 2 standard deviations to the right of mean, correct? So, wouldn't the area under the curve to everything to the left of it be 97.5% or 0.975, according to the Empirical Rule? Why does the z-table show it as 0.9772? I'm genuinely curious.

This is because the 0.975 from the Empirical Rule is just a rough approximation of the probability that the score doesn't exceed two standard deviations above the mean. The 0.9772 you mentioned is indeed a more accurate value of this probability.

something is bothering me on the probability of getting the exact same score of Ludwig (47.5 in this case) by other students. the z-table says 99.38% of students would get less than 47.5 score and the answer for the problem given says 0.62% of them would get higher than 47.5. then must there be 0% of students having 47.5 as their score? statistically yes, but intuitively no. let us picture or imagine this, the test has 50 as full score and 1 problem has 2.5 points. thus any students missing 1 problem would get 47.5 points as their score. even though the probability of a student would get as high as 47.5 like Ludwig's is surely low (around 0.62% as we checked above), it is almost certain that the probability of the event of a student missing 1 problem thus getting 47.5 score is higher than 0 by the design of the test. i guess this might be so clear and easy to someone who is so familiar with the concept of probability. and your answer might be related to the fact that the area of a line must be 0. but that's not persuading me yet. if you have more intuitive or definitive solution, please enlighten me. by the way, thanks for your clear explanation as always Sal

The thing normal distribution is continuous, but the marks that student can take is going to be discrete. Now if you were modelling marks as continuous than yes the probability of getting the mark 47.5 would be 0. But since it is discrete the probability is greater than 0.

Main content

AP®︎/College Statistics

Course: AP®︎/College Statistics > Unit 4

Lesson 5: Normal distribution calculations

Standard normal table for proportion above

Name: Standard normal table for proportion above
Uploaded: 2017-07-12T19:36:04Z
Description: Finding the proportion of a normal distribution that is above a value by calculating a z-score and using a z-table.

AP.STATS:

VAR‑2 (EU)

VAR‑2.B (LO)

VAR‑2.B.3 (EK)

CCSS.Math:

HSS.ID.A.4

Google Classroom

Finding the proportion of a normal distribution that is above a value by calculating a z-score and using a z-table.

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Adam S
Posted 4 years ago. Direct link to Adam S's post “The Unit test says: You m...”
The Unit test says:
You might need: Calculator, Z table

At the bottom is a button: "Show Calculator"
Why is there no button "Show Z Table"?

Considering calculators are quite common both online and physically, I don't think people really need access to a popup calculator, but they certainly would need access to a Z table. So why no link to a Z table?
Button navigates to signup pageComment on Adam S's post “The Unit test says: You m...”
(30 votes)
Answer
- Sky
  Posted 4 months ago. Direct link to Sky's post “I agree, I don’t even use...”
  I agree, I don’t even use the calculator on Khanacademy because of how small it is and just use another website, but a z-table is harder to find and access, and they should have attached it to the assignments and videos.
  Button navigates to signup page
  (1 vote)
sydeegirl00
Posted 4 years ago. Direct link to sydeegirl00's post “What do you do if your z-...”
What do you do if your z-score is negative ?
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(5 votes)
Answer
- Luis Vallejo
  Posted 4 years ago. Direct link to Luis Vallejo's post “Then your value is below ...”
  Then your value is below the mean. Because the table does not have negative values, so assuming that Z follows a normal distribution, you can use P(Z<-x) = 1-P(Z<x)
  Button navigates to signup page
  (5 votes)
Sylvia Wang
Posted 6 years ago. Direct link to Sylvia Wang's post “I now understand how to u...”
I now understand how to use Z table to fix these... but how can we use TI-nspire to calculate the proportion above certain point?
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(8 votes)
Answer
- Raymond Zhang
  Posted 2 months ago. Direct link to Raymond Zhang's post “Use the normalcdf functio...”
  Use the normalcdf function:
  Go to calculator and press
  2nd+vars (distr)
  Then press button 2 (normalcdf)
  
  Depending on your type of TI-calculator, it might give you a different formatting, but the gist is:
  
  normalcdf(lowerbound, upperbound, mean, standard deviation)
  Button navigates to signup page
  (1 vote)
Artur Markhamatov
Posted 4 years ago. Direct link to Artur Markhamatov's post “I'm little confused about...”
I'm little confused about picking up the right values off the Z-Score Table.
Please correct me if I'm wrong about the process of value choosing.
Whenever you figured out the Z-score, you go down the row values, which represent the first two significant figures.
Then you proceed along the column values, that are representing the two decimal places thus making the row Z-score values more accurate.
For example: Suppose Z-score is 2.5 (from the row), then 0.05 (from the column), eventually you are to pick up value where according rows and columns intersect (i.e. z-score=2.55 would give 0.9946?)
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(4 votes)
Answer
- William
  Posted 2 years ago. Direct link to William's post “That seems right, just fi...”
  That seems right, just find the intersection to approximate the value more accurately.
  Button navigates to signup page
  (1 vote)
David Kim
Posted 3 years ago. Direct link to David Kim's post “In one of the practice pr...”
In one of the practice problems, the z-score was 2. This means that it's 2 standard deviations to the right of mean, correct? So, wouldn't the area under the curve to everything to the left of it be 97.5% or 0.975, according to the Empirical Rule? Why does the z-table show it as 0.9772?

I'm genuinely curious.
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- Ian Pulizzotto
  Posted 3 years ago. Direct link to Ian Pulizzotto's post “This is because the 0.975...”
  This is because the 0.975 from the Empirical Rule is just a rough approximation of the probability that the score doesn't exceed two standard deviations above the mean. The 0.9772 you mentioned is indeed a more accurate value of this probability.
  Button navigates to signup page
  (2 votes)
Tessanor
Posted 3 years ago. Direct link to Tessanor's post “how does one come up with...”
how does one come up with these proportions without the table?
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(2 votes)
Answer
- Raymond Zhang
  Posted 2 months ago. Direct link to Raymond Zhang's post “Use the normalcdf functio...”
  Use the normalcdf function:
  Go to calculator and press
  2nd+vars (distr)
  Then press button 2 (normalcdf)
  
  Depending on your type of TI-calculator, it might give you a different formatting, but the gist is:
  
  normalcdf(lowerbound, upperbound, mean, standard deviation)
  Button navigates to signup page
  (1 vote)
deka
Posted 10 months ago. Direct link to deka's post “something is bothering me...”
something is bothering me on the probability of getting the exact same score of Ludwig (47.5 in this case) by other students.

the z-table says 99.38% of students would get less than 47.5 score and the answer for the problem given says 0.62% of them would get higher than 47.5.

then must there be 0% of students having 47.5 as their score?
statistically yes, but intuitively no.

let us picture or imagine this, the test has 50 as full score and 1 problem has 2.5 points. thus any students missing 1 problem would get 47.5 points as their score. even though the probability of a student would get as high as 47.5 like Ludwig's is surely low (around 0.62% as we checked above), it is almost certain that the probability of the event of a student missing 1 problem thus getting 47.5 score is higher than 0 by the design of the test.

i guess this might be so clear and easy to someone who is so familiar with the concept of probability. and your answer might be related to the fact that the area of a line must be 0. but that's not persuading me yet. if you have more intuitive or definitive solution, please enlighten me.

by the way, thanks for your clear explanation as always Sal
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- cossine
  Posted 10 months ago. Direct link to cossine's post “The thing normal distribu...”
  The thing normal distribution is continuous, but the marks that student can take is going to be discrete. Now if you were modelling marks as continuous than yes the probability of getting the mark 47.5 would be 0. But since it is discrete the probability is greater than 0.
  Button navigates to signup page
  (2 votes)
Yevheniia Rakhmatova
Posted 5 years ago. Direct link to Yevheniia Rakhmatova's post “Where do I get z-scores t...”
Where do I get z-scores table?
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(1 vote)
Answer
- Nathan
  Posted 4 years ago. Direct link to Nathan's post “All or at least most z-ta...”
  All or at least most z-tables are similar and can be found by performing a simple Google Search. All are fairly alike because a z-table corresponds to the area under a bell curve.
  Button navigates to signup page
  (1 vote)
busyargirl
Posted 7 months ago. Direct link to busyargirl's post “Are normal distributions ...”
Are normal distributions abstract? In real life could you actually have a perfectly normal distribution?
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(1 vote)
Answer
ANB
Posted a year ago. Direct link to ANB's post “What z-table is Sal using...”
What z-table is Sal using? It seems like there are minor variations between different z-tables and that is leading to my answer getting marked wrong.
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(1 vote)
Answer
- Raymond Zhang
  Posted 2 months ago. Direct link to Raymond Zhang's post “He is using the official ...”
  He is using the official AP Stats stat-table
  Button navigates to signup page
  (1 vote)

Video transcript

- [Tutor] A set of philosophy exam scores are normally distributed with a mean of 40 points and a standard deviation of three points. Ludwig got a score of 47.5 points on the exam. What proportion of exam scores are higher than Ludwig's score? Give your answer correct to four decimal places. So, let's just visualize what's going on here. So, the scores are normally distributed. So, it would look like this. So, the distribution would look something like that, trying to make that pretty symmetric looking. The mean is 40 points, so that would be 40 points right over there. Standard deviation is three points, so this could be one standard deviation above the mean, that would be one standard deviation below the mean. And once again, this is just very rough. And so, this would be 43, this would be 37 right over here. And they say Ludwig got a score of 47.5 points on the exam. So, Ludwig's score is going to be someplace around here. So, Ludwig got a 47.5 on the exam. And they're saying, what proportion of exam scores are higher than Ludwig's score? So, what we need to do is figure out what is the area under the normal distribution curve that is above 47.5. So, the way we will tackle this is we will figure out the z score for 47.5. How many standard deviations above the mean is that? Then, we will look at a z table to figure out what proportion is below that because that's what z tables give us. They give us the proportion that is below a certain z score. And then, we can take one minus that to figure out the proportion that is above. Remember, the entire area under the curve is one, so if we can figure out this orange area and take one minus that, we're gonna get the red area. So, let's do that. So, first of all, let's figure out the z score for 47.5. So, let's see. We would take 47.5 and we would subtract the mean. So, this is his score. We'll subtract the mean, minus 40. We know what that's gonna be, that's 7.5. That's how much more above the mean. But how many standard deviations is that? Well, each standard deviation is three, so what's 7.5 divided by three? This just means the previous answer divided by three. So, he is 2.5 standard deviations above the mean. So, the z score here, z score here is a positive 2.5. If he was below the mean, it would be a negative. So now, we can look at a z table to figure out what proportion is less than 2.5 standard deviations above the mean. So, that'll give us that orange and then we'll subtract that from one. So, let's get our z table. So, here we go. And we've already done this in previous videos, but what's going on here is this left column gives us our z score up to the tenths place. And then these other columns give us the hundredths place. So, what we want to do is find 2.5 right over here on the left, and it's actually gonna be 2.50. There's zero hundredths here. So, we want to look up 2.50. Let me scroll my z table. So, I'm gonna go down to 2.5. Alright, I think I am there. So, what I have here, so I have 2.5, so I am going to be in this row. And it's now scrolled off, but this first column we saw, this is the hundredths place and this zero hundredths. And so, 2.50 puts us right over here. Now, you might be tempted to say, okay, that's the proportion that scores higher than Ludwig, but you'd be wrong. This is the proportion that scores lower than Ludwig. So, what we wanna do is take one minus this value. So, let me get my calculator out again. So, what I'm going to do is I'm going to take one minus this. One minus 0.9938 is equal to, now this is, so this is the proportion that scores less than Ludwig. One minus that is gonna be the proportion that scores more than him. The reason why we have to do this is because the z table gives us the proportion less than a certain z score. So, this gives us right over here, 0.0062. So, that's the proportion. If you thought of it in percent, it would be 0.62% scores higher than Ludwig. Now, that makes sense 'cause Ludwig scored over two standard deviations, two and a half standard deviations above the mean. So, our answer is 0.0062. So, this is going to be 0.0062. That's the proportion of exam scores higher than Ludwig's score.