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

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.

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

Method 1: Using the standard normal table

Write a probability statement for $Z$

$P\mathit{(}\mathit{Z}\mathit{<}\mathit{z}\mathit{)}=0.1$

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

Find $P\left(Z>z\right)=0.1$

$P\left(Z>z\right)=0.1\Rightarrow z=1.282$

Change the sign as the value is below the mean

$z=-1.282$

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

$\begin{array}{rcl}-1.282& =& \frac{12-\mu }{4}\\ -1.282·4& =& 12-\mu \\ \mu & =& 12+1.282·4\\ \mu & =& 17.128\end{array}$

Explain the value in the context of the question

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

Method 2: Using a calculator

Write a probability statement for $Z$

$P\mathit{(}\mathit{Z}\mathit{<}\mathit{z}\mathit{)}=0.1$

Write down the parameters of the situation

probability, $p$, (area) = 0.1

$\mu =0$

$\sigma =1$

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 $z=\frac{x-\mu }{\sigma }$

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

Explain the value in the context of the question

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

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.

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

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 $H$

Write the corresponding probability statements for $Z$

$$\begin{gathered} P\mathit{(}Z\mathit{<}{z}_{\mathit{2}\mathit{.}\mathit{2}}\mathit{)}=0.15 \\ P\mathit{(}Z>z_{2.8}\mathit{)}=0.3 \end{gathered}$$

Convert the second statement to a 'less than' probability

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

Method 1: Using the tables

$z_{2.2}$ can be found using the ∞ row in the table for t-distribution critical values

$P\left(Z>z\right)=0.15\Rightarrow z=1.036$

Change the sign as the value is below the mean

$z_{2.2}=-1.036$

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

$\begin{array}{rcl}-1.036& =& \frac{2.2-\mu }{\sigma }\end{array}$

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 $z=\frac{x-\mu }{\sigma }$

$\begin{array}{rcl}0.53& =& \frac{2.8-\mu }{\sigma }\end{array}$

Rearrange both equations

$\begin{array}{rcl}-1.036& =& \frac{2.2-\mu }{\sigma }\\ 2.2& =& \mu -1.036\sigma \end{array}$ and $\begin{array}{rcl}0.53& =& \frac{2.8-\mu }{\sigma }\\ 2.8& =& \mu +0.53\sigma \end{array}$

Solve simultaneously using your calculator

$$\begin{gathered} \mu =2.5969...\approx 2.60 \\ \sigma =0.3831...\approx 0.383 \end{gathered}$$

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 $z_{2.2}$

probability, $p$, (area) = 0.15

$\mu =0$

$\sigma =1$

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 $z=\frac{x-\mu }{\sigma }$

$\begin{array}{rcl}-1.03643...& =& \frac{2.2-\mu }{\sigma }\end{array}$

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

probability, $p$, (area) = 0.7

$\mu =0$

$\sigma =1$

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 $z=\frac{x-\mu }{\sigma }$

$\begin{array}{rcl}0.52440...& =& \frac{2.8-\mu }{\sigma }\end{array}$

Rearrange both equations

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

Solve simultaneously

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

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