Estimating Parameters of Normal Distributions (College Board AP® Statistics): Study Guide

Syllabus Edition

First teaching 2026

First exams 2027

Naomi C

Written by: Naomi C

Reviewed by: Dan Finlay

Updated on

Estimating one parameter of a normal distribution

How do I find the mean (μ) or the standard deviation (σ) if one of them is unknown using the standard normal table?

  • If the mean or standard deviation of a normal distribution of a variable X is unknown then you will need to use the standard normal distribution

  • You will be given a probability for a specific value of x 

    • P left parenthesis X less than x right parenthesis equals p or P left parenthesis X greater than x right parenthesis equals p

    • You will need P left parenthesis X less than x right parenthesis equals p for the tables, so remember P open parentheses X greater than x close parentheses equals 1 minus P open parentheses X less than x close parentheses

  • To find the unknown parameter using the standard normal table:

    • sketch the normal curve

      • label the given value, x

      • and the known parameter, either mu or sigma

    • use the standard normal table to identify the z-score for the given value, x

      • remember that the table lists the proportions (or probabilities) that are less than a given value

      • so make sure that straight P open parentheses Z less than z close parentheses equals p

    • substitute the known values into z equals fraction numerator x minus mu over denominator sigma end fraction

      • you have calculated z

      • and you know both x and either muor sigma

    • solve the equation to find the unknown parameter, mu or sigma

How do I find the mean (μ) or the standard deviation (σ) if one of them is unknown using a calculator?

  • If using a calculator to find the unknown parameter:

    • sketch the normal curve

      • label the given value, x

      • and the known parameter, either mu or sigma

    • use the Inverse Normal Distribution on your calculator to find the z-score of x

      • if straight P left parenthesis Z less than z right parenthesis equals p, enter:

        • the proportion (or probability), p

        • the mean for the standard normal distribution, mu equals 0

        • the standard deviation for the standard normal distribution, sigma equals 1

      • or if straight P left parenthesis Z greater than z right parenthesis equals p, find straight P left parenthesis Z less than z right parenthesis equals 1 minus p first

    • then substitute the known values into z equals fraction numerator x minus mu over denominator sigma end fraction and solve for the unknown parameter as before

  • Some calculators might ask for the tail, you can use the given probability P left parenthesis Z less than z right parenthesis equals p or P left parenthesis Z greater than z right parenthesis equals p, and select the correct tail

    • For P open parentheses Z less than z close parentheses this is the left tail

    • For open parentheses Z greater than z close parentheses this is the right tail

Examiner Tips and Tricks

Be careful to avoid rounding errors.

When taking the z-score from a calculator use at least one extra decimal place within your working than your intended degree of accuracy for your answer.

Worked Example

It is known that the times, in minutes, taken by students at a school to eat their lunch is approximately normal with a standard deviation of 4 minutes. Given that 10 percent of students at the school take less than 12 minutes to eat their lunch, what is the mean time taken to eat lunch by the students at the school?

Answer:

Define your variable and its distribution

Let T be the time taken for a student to eat their lunch

The distribution of T is approximately normal with mean mu and standard deviation 4

Draw a sketch describing the proportion of the distribution you are trying to find and write a probability statement for T

Even though we don't know the mean of the distribution, we do know that 10% of students take less than 12 minutes, so 12 must lie below the mean

Normal distribution with mean 𝞵 and an area shaded to the left of t=12. Shaded area is labeled P(T<12)=0.1.

Method 1: Using the standard normal table

Write a probability statement for Z

Error converting from MathML to accessible text.

You can use the ∞ row in the table for t-distribution critical values

Find P open parentheses Z greater than z close parentheses equals 0.1

P open parentheses Z greater than z close parentheses equals 0.1 rightwards double arrow z equals 1.282

Change the sign as the value is below the mean

z equals negative 1.282

Convert the z-score into an actual value, using z equals fraction numerator x minus mu over denominator sigma end fraction

table row cell negative 1.282 end cell equals cell fraction numerator 12 minus mu over denominator 4 end fraction end cell row cell negative 1.282 times 4 end cell equals cell 12 minus mu end cell row mu equals cell 12 plus 1.282 times 4 end cell row mu equals cell 17.128 end cell end table

Explain the value in the context of the question

The mean time taken for a student to eat their lunch is 17.1 minutes

Method 2: Using a calculator

Write a probability statement for Z

Error converting from MathML to accessible text.

Write down the parameters of the situation

probability, p, (area) = 0.1

mu equals 0

sigma equals 1

Enter these values into the Inverse Normal Distribution function on your calculator

table row z equals cell negative 1.281551... end cell end table

Convert the z-score into an actual value, using z equals fraction numerator x minus mu over denominator sigma end fraction

table row cell negative 1.281551... end cell equals cell fraction numerator 12 minus mu over denominator 4 end fraction end cell row cell negative 1.281551... times 4 end cell equals cell 12 minus mu end cell row mu equals cell 12 plus 1.281551... times 4 end cell row mu equals cell 17.126204... end cell row mu almost equal to cell 17.1 end cell end table

Explain the value in the context of the question

The mean time taken for a student to eat their lunch is 17.1 minutes

Examiner Tips and Tricks

You may get slightly different answers depending on whether you have used the standard normal table or a calculator, as the z-score will be given to different levels of accuracy.

As long as your method is correct and full working is shown, your answer will be accepted.

Estimating both parameters of a normal distribution

How do I find the mean (μ) and the standard deviation (σ) if both of them are unknown?

  • If both the mean and the standard deviation of a normal distribution are unknown, you will be given two proportions (or probabilities) for two specific values of x

    • These can be used to form a pair of simultaneous equations that you can solve to find both parameters

  • The process is the same as above for both the standard normal table and the calculator

    • calculate two z-scores

    • form two equations using z equals fraction numerator x minus mu over denominator sigma end fraction

      • rearranging to x equals mu plus sigma z can be helpful

    • solve the two equations simultaneously

      • you can use your calculator to check your solution

Examiner Tips and Tricks

When working with more than one proportion (or probability) be careful not to confuse which z-score goes with which probability!

Worked Example

The distribution of the heights of sunflowers from a garden center follows approximately a normal distribution. Based on a very large sample, it was found that 15 percent had a height of less than 2.2 meters, and 30 percent of the sunflowers had a height of more than 2.8 meters. What are the mean and standard deviation of the distribution of the heights of the sunflowers?

Answer:

Define your variable and its distribution

Let H be the height of a sunflower

The distribution of H is approximately normal with mean mu and standard deviation sigma

Draw a sketch describing the proportion of the distribution you are trying to find and write two probability statements for H

Normal distribution curve with mean μ; shaded left tail for P(H < 2.2) = 0.15 and shaded right tail for P(H > 2.8) = 0.3, with 2.2 and 2.8 marked.

Write the corresponding probability statements for Z

Error converting from MathML to accessible text.

Convert the second statement to a 'less than' probability

Error converting from MathML to accessible text.

Method 1: Using the tables

z subscript 2.2 end subscript can be found using the ∞ row in the table for t-distribution critical values

P open parentheses Z greater than z close parentheses equals 0.15 rightwards double arrow z equals 1.036

Change the sign as the value is below the mean

z subscript 2.2 end subscript equals negative 1.036

Substitute the z-score and h equals 2.2 into z equals fraction numerator x minus mu over denominator sigma end fraction

table row cell negative 1.036 end cell equals cell fraction numerator 2.2 minus mu over denominator sigma end fraction end cell end table

Using the standard normal table, look at section with the positive z-values for z subscript 2.8 end subscript

Find the cell that contains the probability, 0.7, if there is no exact value, find the smallest value that is greater than 0.7

0.7019 is the smallest value in the standard normal table that is greater than 0.7

Write down the z-score from the corresponding row and column

z equals 0.53

Substitute the z-score and h equals 2.8 into z equals fraction numerator x minus mu over denominator sigma end fraction

table row cell 0.53 end cell equals cell fraction numerator 2.8 minus mu over denominator sigma end fraction end cell end table

Rearrange both equations

table row cell negative 1.036 end cell equals cell fraction numerator 2.2 minus mu over denominator sigma end fraction end cell row cell 2.2 end cell equals cell mu minus 1.036 sigma end cell end table and table row cell 0.53 end cell equals cell fraction numerator 2.8 minus mu over denominator sigma end fraction end cell row cell 2.8 end cell equals cell mu plus 0.53 sigma end cell end table

Solve simultaneously using your calculator

mu equals 2.5969... almost equal to 2.60
sigma equals 0.3831... almost equal to 0.383

Explain the values in the context of the question

The mean height of the sunflowers is 2.60 meters and the standard deviation is 0.383 meters

Method 2: Using a calculator

Write down the parameters of the situation for z subscript 2.2 end subscript

probability, p, (area) = 0.15

mu equals 0

sigma equals 1

Enter these values into the Inverse Normal Distribution function on your calculator

table row cell z subscript 2.2 end subscript end cell equals cell negative 1.03643... end cell end table

Substitute the z-score and h equals 2.2 into z equals fraction numerator x minus mu over denominator sigma end fraction

table row cell negative 1.03643... end cell equals cell fraction numerator 2.2 minus mu over denominator sigma end fraction end cell end table

Write down the parameters of the situation for z subscript 2.8 end subscript

probability, p, (area) = 0.7

mu equals 0

sigma equals 1

Enter these values into the Inverse Normal Distribution function on your calculator

table row cell z subscript 2.8 end subscript end cell equals cell 0.52440... end cell end table

Substitute the z-score and h equals 2.8 into z equals fraction numerator x minus mu over denominator sigma end fraction

table row cell 0.52440... end cell equals cell fraction numerator 2.8 minus mu over denominator sigma end fraction end cell end table

Rearrange both equations

table row cell negative 1.03643... end cell equals cell fraction numerator 2.2 minus mu over denominator sigma end fraction end cell row sigma equals cell fraction numerator 2.2 minus mu over denominator negative 1.03643... end fraction end cell end table and table row cell 0.52440... end cell equals cell fraction numerator 2.8 minus mu over denominator sigma end fraction end cell row sigma equals cell fraction numerator 2.8 minus mu over denominator 0.52440... end fraction end cell end table

Solve simultaneously

table row sigma equals cell fraction numerator 2.2 minus 2.59841... over denominator negative 1.03643... end fraction end cell row sigma equals cell 0.38441... end cell end table

Explain the values in the context of the question

The mean height of the sunflowers is 2.60 meters and the standard deviation is 0.384 meters

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Naomi C

Author: Naomi C

Expertise: Maths Content Creator

Naomi graduated from Durham University in 2007 with a Masters degree in Civil Engineering. She has taught Mathematics in the UK, Malaysia and Switzerland covering GCSE, IGCSE, A-Level and IB. She particularly enjoys applying Mathematics to real life and endeavours to bring creativity to the content she creates.

Dan Finlay

Reviewer: Dan Finlay

Expertise: Maths Subject Lead

Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.