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### Course: Statistics and probability>Unit 4

Lesson 6: Normal distribution calculations

# Standard normal table for proportion between values

Finding the proportion of a normal distribution that is between two values by calculating z-scores and using a z-table.

## Want to join the conversation?

• The z-table I get to use in class doesn't go below 0, how do I deal with this?
• Here is a link to Z-Tables for Z-Scores that are negative and positive.
http://www.math.odu.edu/stat130/normal-tables.pdf

Or you can take the value on the z-Table of the positive Z-Score and subtract it from 1.
Example: Z-Score = -1.20, Z-Table for +1.20 = 0.8849
1-0.8849 = 0.1151. That is the same as if you look at the Z-Table for values with negative Z-Scores.
• Hi Sal,

I am curious about how to find the area under the Normal curve between two values without using a table or a calculator. If you think it would be useful, you could make a video with an example of a definite integral calculation for a Normal curve and put it in the "Applications of integration" unit in AP Calculus BC.

Thank you for reading this! It is very much appreciated. :)
• At -, how do we already know it will be 2 point something?
• Z-scores are measuring the number of standard deviations a certain number is above or below the mean. At , we learn that 750(the mean) - 60(the standard deviation) is 690. If you take another standard deviation away from that, it will equal 630. 624 is more than 120(2 standard deviations) below 750, and, therefore, its z-score will -2 point something.
• My answer became 0.3642, as I subtracted the z-score of 624, which is 0.0179, from 1 and obtained 0.9821. After that, I further subtracted 0.9821 by 0.6179, resulting in 0.3642.
• Looking at your reasoning, you got:
- The % of laptops above 624
- The % of laptops below 768

Subtracting the z-score of 0.6179 from your score of 0.9821, you get the % of laptops that are above 624, but not below 768; you essentially only get the % of laptops above 768.

What you would want to do is, instead, subtract your 624 z-score (which is 0.01786) from your 768 z-score (which is 0.61791) to get the % of laptops that are below 768, but not below 624, which is what we want. You would end up with a score of 0.60005, or ~60%.
• Shouldn't this have been calculated "higher than 624" and "less than 768"; Eqn being {Zscrore768 - (1 - Zscore624)}?
(1 vote)
• So you would need to use the cdf: culmative density function of the standard normal distribution to calculate the probability.

Observe P(624<X<768) = P(X<768) - P(X<624) .

After you calculate the z-score you can apply the cdf given by the z-table.
• Is there a way to get the Z-table values mathematically?

If so, I would love to know!
(1 vote)
• z
∫ (1 / √(2π)) * e⁻⁽ˣ^²⁾ᐟ²
-∞
• how do i find the range of values if i have the percentage?
(1 vote)
• you can't, because RANGE is determined by 2 variables, upper boundary and lower boundary
but if you know one of the boundaries, you can use the integral and the anti-derivative(the second fundamental theorem of calculus, if my memory doesn't fool me) to get the other bound
• Is there another solution to solve the problem than having to use a z-table?
(1 vote)
• yes, we have, with the help of integral.
1. find the general PDF for normal distribution from https://www.itl.nist.gov/div898/handbook/eda/section3/eda3661.htm
2. fill in the parameters, μ and σ with 750 and 60 respectively, we got the PDF for this distribution
3. integrate the result from 624 to 768, we got the same result(to evaluate it, you need a calculator), 0.600047001626
``import scipy.stats as stimport numpy as npmean = 19.7stdev = 2datapoint = 21.4zscore = np.round((datapoint-mean)/stdev, 4)# SciPy calculates left-tail probabilities by defaultauc = np.round(st.norm.cdf(zscore), 4)print(f"Z-score: {zscore}\tArea under curve: {auc}\tRemainder: {1-auc}")``