If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Standard normal table for proportion above

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

Want to join the conversation?

  • leaf green style avatar for user Adam S
    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?
    (42 votes)
    Default Khan Academy avatar avatar for user
    • female robot grace style avatar for user Sky
      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.
      (12 votes)
  • blobby green style avatar for user sydeegirl00
    What do you do if your z-score is negative ?
    (5 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user Sylvia Wang
    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?
    (8 votes)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user Raymond Zhang
      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)
      (2 votes)
  • blobby green style avatar for user David Kim
    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.
    (3 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user Artur Markhamatov
    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?)
    (4 votes)
    Default Khan Academy avatar avatar for user
  • piceratops ultimate style avatar for user ANB
    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.
    (3 votes)
    Default Khan Academy avatar avatar for user
  • winston default style avatar for user Lora
    Where is the z-table on the exercises?
    (4 votes)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user daniella
      The z-table is typically provided separately as a reference tool in textbooks or online resources. In exercises or exams, you may be expected to use a z-table to find probabilities associated with specific z-scores. If the z-table is not provided, you can use a calculator or statistical software to compute the probabilities.
      (1 vote)
  • purple pi pink style avatar for user Tessanor
    how does one come up with these proportions without the table?
    (3 votes)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user Raymond Zhang
      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)
      (2 votes)
  • blobby green style avatar for user LinkenR
    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 video
    (3 votes)
    Default Khan Academy avatar avatar for user
  • starky seedling style avatar for user deka
    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
    (1 vote)
    Default Khan Academy avatar avatar for user
    • leaf green style avatar for user cossine
      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.
      (4 votes)

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.