Question

The boxplots below display annual incomes (in thousands of dollars) for households in two cities. Which city has a greater percentage of households with annual incomes between 50,000 dollars and 80,000 dollars? Statstown Medianville Both cities have the same percentage of households with annual incomes between 50,000 dollars and 80,000 dollars. It is impossible to tell from the boxplots.

Answer 1

Explanation:

Hello!

Box-plots in the attachment.

Both graphs show information about the annual incomes of households in "Statstown" and "Medianville".

To compare both populations and determine which one has a greater percentage of households with the annual income you have to compare the boxes of both diagrams.

Remember the box is determined by the values of the first (or lower) Quartile -Q1- and the third (or upper) Quartile -Q3-. Between these two values, you find the middle 50% of the data of the sample and the distance between them is the IQR.

The greater the IQR is, the longer is the box and this means that the data in the middle of the distribution is more dispersed.

Looking at the box of the graphic of "Statstown" you can see that Q1= 40 and Q3 is a little more than 100, for example, Q3 ≅ 105.

The second Quantile (or Median) -Q2- is 80

Now the box of the graphic of "Medianville"

Q1= 60

Q3 is equal as the one for "Statstown" so let's say Q3 ≅ 105

Q2= 80

You have to see which one has a larger percentage of the sample between 50 and 80.

Considering that between Q1 and Q2 is 25%

For "Statstown" you can say that there is 25% between Q1= 40 and Q2=80, this means that there will be less than 25% between 50 and 80.

For "Medianville" there is 25% of the sample between Q1=60 and Q2=80, looking at the graphic this means that there is more than 25% between 50 and 80.

"Medianville" has a greater percentage of households with annual income between $50000 and $80000 than "Statstown"

I hope it helps!