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# 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.

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• 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?
• 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.
• 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)
• 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?
• 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)
• 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.
• 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?)
• That seems right, just find the intersection to approximate the value more accurately.
• 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.
• He is using the official AP Stats stat-table
• Where is the z-table on the exercises?
• 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)
• how does one come up with these proportions without the table?
• 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)
• 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
• open z-score.net website
(1 vote)
• 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)
• 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.