Estimating Parameters of Normal Distributions (College Board AP® Statistics)
Study Guide
Written by: Naomi C
Reviewed by: Dan Finlay
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 is unknown then you will need to use the standard normal distribution
You will be given a probability for a specific value of
or
You will need for the tables, so remember
To find the unknown parameter using the standard normal table:
sketch the normal curve
label the given value,
and the known parameter, either or
use the standard normal table to identify the -score for the given value,
remember that the table lists the proportions (or probabilities) that are less than a given value
so make sure that
substitute the known values into
you have calculated
and you know both and either or
solve the equation to find the unknown parameter, or
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,
and the known parameter, either or
use the Inverse Normal Distribution on your calculator to find the -score of
if , enter:
the proportion (or probability),
the mean for the standard normal distribution,
the standard deviation for the standard normal distribution,
or if , find first
then substitute the known values into and solve for the unknown parameter as before
Some calculators might ask for the tail, you can use the given probability or , and select the correct tail
For this is the left tail
For this is the right tail
Examiner Tips and Tricks
Be careful to avoid rounding errors.
When taking the -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 of4 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:
Method 1: Using the standard normal table
Define your variable and its distribution
Let be the time taken for a student to eat their lunch
The distribution of is approximately normal with mean and standard deviation 4
Draw a sketch describing the proportion of the distribution you are trying to find and write a probability statement for
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
Write a probability statement for
Using the standard normal table, look at section with the negative -values as the upper bound of the region is below the mean
The standard normal table contains the proportions or percentages that are below a given value, i.e. to the left of the given value
Find the cell that contains the given probability, 0.1, if there is no exact value, find the greatest value that is smaller than 0.1
0.0985 is the greatest value in the standard normal table that is smaller than 0.1
Write down the -score from the corresponding row and column
Convert the -score into an actual value, using
Explain the value in the context of the question
The mean time taken for a student to eat their lunch is 17.2 minutes
Method 2: Using a calculator
Define your variable and its distribution
Let be the time taken for a student to eat their lunch
The distribution of is approximately normal with mean and standard deviation 4
Draw a sketch describing the proportion of the distribution you are trying to find and write a probability statement for
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
Write a probability statement for
Write down the parameters of the situation
probability, , (area) = 0.1
Enter these values into the Inverse Normal Distribution function on your calculator
Convert the -score into an actual value, using
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 -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
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
rearranging to 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 -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 height of more than 2.8 meters. What are the mean and standard deviation of the distribution of the heights of the sunflowers?
Answer:
Method 1: Using the standard normal table
Define your variable and its distribution
Let be the height of a sunflower
The distribution of is approximately normal with mean and standard deviation
Draw a sketch describing the proportion of the distribution you are trying to find and write two probability statements for
Write the corresponding probability statements for
Convert the second statement to a 'less than' probability
Using the standard normal table, look at section with the negative -values for
Find the cell that contains the given probability, 0.15, if there is no exact value, find the greatest value that is smaller than 0.1
0.1492 is the greatest value in the standard normal table that is smaller than 0.1
Write down the -score from the corresponding row and column
Substitute the -score and into
Using the standard normal table, look at section with the positive -values for
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 -score from the corresponding row and column
Substitute the -score and into
Rearrange both equations
and
Solve simultaneously using your calculator
Explain the values in the context of the question
The mean time height of the sunflowers is 2.60 meters and the standard deviation is 0.382 meters
Method 2: Using a calculator
Define your variable and its distribution
Let be the height of a sunflower
The distribution of is approximately normal with mean and standard deviation
Draw a sketch describing the proportion of the distribution you are trying to find and write two probability statements for
Write the corresponding probability statements for
Convert the second statement to a 'less than' probability
Write down the parameters of the situation for
probability, , (area) = 0.15
Enter these values into the Inverse Normal Distribution function on your calculator
Substitute the -score and into
Write down the parameters of the situation for
probability, , (area) = 0.7
Enter these values into the Inverse Normal Distribution function on your calculator
Substitute the -score and into
Rearrange both equations
and
Solve simultaneously (you can check your solutions using your calculator)
Explain the values in the context of the question
The mean time height of the sunflowers is 2.60 meters and the standard deviation is 0.384 meters
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