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

Revision Note

Author

Naomi C

Expertise

Maths

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 $x$
- $P (X < x) = p$ or $P (X > x) = p$
- You will need $P (X < x) = p$ for the tables, so remember $P (X > x) = 1 - P (X < x)$
To find the unknown parameter using the standard normal table:
- sketch the normal curve
  - label the given value, $x$
  - and the known parameter, either $μ$ or $σ$
- 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 $P (Z < z) = p$
- substitute the known values into $z = \frac{x - μ}{σ}$
  - you have calculated $z$
  - and you know both $x$ and either $μ$ or $σ$
- solve the equation to find the unknown parameter, $μ$ or $σ$

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 $μ$ or $σ$
- use the Inverse Normal Distribution on your calculator to find the $z$ -score of $x$
  - if $P (Z < z) = p$ , enter:
    - the proportion (or probability), $p$
    - the mean for the standard normal distribution, $μ = 0$
    - the standard deviation for the standard normal distribution, $σ = 1$
  - or if $P (Z > z) = p$ , find $P (Z < z) = 1 - p$ first
- then substitute the known values into $z = \frac{x - μ}{σ}$ and solve for the unknown parameter as before
Some calculators might ask for the tail, you can use the given probability $P (Z < z) = p$ or $P (Z > z) = p$ , and select the correct tail
- For $P (Z < z)$ this is the left tail
- For $(Z > z)$ this is the right tail

Exam Tip

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

The distribution of $T$ is approximately normal with mean $μ$ 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.

Write a probability statement for $Z$

$P (Z < z) = 0.1$

Using the standard normal table, look at section with the negative $z$ -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 $z$ -score from the corresponding row and column

$z = - 1.29$

Convert the $z$ -score into an actual value, using $z = \frac{x - μ}{σ}$

$\begin{array}{rcl} - 1.29 & = & \frac{12 - μ}{4} \\ - 1.29 \cdot 4 & = & 12 - μ \\ μ & = & 12 + 1.29 \cdot 4 \\ μ & = & 17.16 \end{array}$

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

The distribution of $T$ is approximately normal with mean $μ$ 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

Write a probability statement for $Z$

$P (Z < z) = 0.1$

Write down the parameters of the situation

probability, $p$ , (area) = 0.1

$μ = 0$

$σ = 1$

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

$\begin{array}{rcl} z & = & - 1.281551 . . . \end{array}$

Convert the $z$ -score into an actual value, using $z = \frac{x - μ}{σ}$

$\begin{array}{rcl} - 1.281551 . . . & = & \frac{12 - μ}{4} \\ - 1.281551 . . . \cdot 4 & = & 12 - μ \\ μ & = & 12 + 1.281551 . . . \cdot 4 \\ μ & = & 17.126204 . . . \\ μ & \approx & 17.1 \end{array}$

Explain the value in the context of the question

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

Exam Tip

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 $z = \frac{x - μ}{σ}$
  - rearranging to $x = μ + σ z$ can be helpful
- solve the two equations simultaneously
  - you can use your calculator to check your solution

Exam Tip

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 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 $H$ be the height of a sunflower

The distribution of $H$ is approximately normal with mean $μ$ and standard deviation $σ$

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

Write the corresponding probability statements for $Z$

$P (Z < z_{2.2}) = 0.15 P (Z > z_{2.8}) = 0.3$

Convert the second statement to a 'less than' probability

$\begin{array}{rcl} P (Z < z_{2.8}) & = & 1 - P (Z > z_{2.8}) \\ = & 1 - 0.3 \\ = & 0.7 \end{array}$

Using the standard normal table, look at section with the negative $z$ -values for $z_{2.2}$

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 $z$ -score from the corresponding row and column

$z = - 1.04$

Substitute the $z$ -score and $h = 2.2$ into $z = \frac{x - μ}{σ}$

$\begin{array}{rcl} - 1.04 & = & \frac{2.2 - μ}{σ} \end{array}$

Using the standard normal table, look at section with the positive $z$ -values for $z_{2.8}$

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

$z = 0.53$

Substitute the $z$ -score and $h = 2.8$ into $z = \frac{x - μ}{σ}$

$\begin{array}{rcl} 0.53 & = & \frac{2.8 - μ}{σ} \end{array}$

Rearrange both equations

$\begin{array}{rcl} - 1.04 & = & \frac{2.2 - μ}{σ} \\ 2.2 & = & μ - 1.04 σ \end{array}$ and $\begin{array}{rcl} 0.53 & = & \frac{2.8 - μ}{σ} \\ 2.8 & = & μ + 0.53 σ \end{array}$

Solve simultaneously using your calculator

$μ = 2.59745 . . . \approx 2.60 σ = 0.38216 . . . \approx 0.382$

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 $H$ be the height of a sunflower

The distribution of $H$ is approximately normal with mean $μ$ and standard deviation $σ$

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

Write the corresponding probability statements for $Z$

$P (Z < z_{2.2}) = 0.15 P (Z > z_{2.8}) = 0.3$

Convert the second statement to a 'less than' probability

$\begin{array}{rcl} P (Z < z_{2.8}) & = & 1 - P (Z > z_{2.8}) \\ = & 1 - 0.3 \\ = & 0.7 \end{array}$

Write down the parameters of the situation for $z_{2.2}$

probability, $p$ , (area) = 0.15

$μ = 0$

$σ = 1$

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

$\begin{array}{rcl} z_{2.2} & = & - 1.03643 . . . \end{array}$

Substitute the $z$ -score and $h = 2.2$ into $z = \frac{x - μ}{σ}$

$\begin{array}{rcl} - 1.03643 . . . & = & \frac{2.2 - μ}{σ} \end{array}$

Write down the parameters of the situation for $z_{2.8}$

probability, $p$ , (area) = 0.7

$μ = 0$

$σ = 1$

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

$\begin{array}{rcl} z_{2.8} & = & 0.52440 . . . \end{array}$

Substitute the $z$ -score and $h = 2.8$ into $z = \frac{x - μ}{σ}$

$\begin{array}{rcl} 0.52440 . . . & = & \frac{2.8 - μ}{σ} \end{array}$

Rearrange both equations

$\begin{array}{rcl} - 1.03643 . . . & = & \frac{2.2 - μ}{σ} \\ σ & = & \frac{2.2 - μ}{- 1.03643 . . .} \end{array}$ and $\begin{array}{rcl} 0.52440 . . . & = & \frac{2.8 - μ}{σ} \\ σ & = & \frac{2.8 - μ}{0.52440 . . .} \end{array}$

Solve simultaneously (you can check your solutions using your calculator)

$\begin{array}{rcl} \frac{2.2 - μ}{- 1.03643 . . .} & = & \frac{2.8 - μ}{0.52440 . . .} \\ 0.52440 . . . \cdot 2.2 - 0.52440 . . . μ & = & - 1.03643 . . . \cdot 2.8 + 1.03643 . . . μ \\ 4.05568 . . . & = & 1.56083 . . . μ \\ μ & = & 2.59841 . . . \end{array}$

$\begin{array}{rcl} σ & = & \frac{2.2 - 2.59841 . . .}{- 1.03643 . . .} \\ σ & = & 0.38441 . . . \end{array}$

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

You've read 0 of your 10 free revision notes

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

Next topic

Did this page help you?

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

Revision Note

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?

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

Exam Tip

Worked Example

Exam Tip

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?

Exam Tip

Worked Example

You've read 0 of your 10 free revision notes

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

Author: Naomi C