Question

By how many units is the interquartile range for Orlando's data set greater than the interquartile range for Victor's data set? Victor's Data Set: {22, 29, 32, 27, 30} Orlando's Data Set: {36, 25, 33, 27, 35}

Answer 1

Answer:

By 3 units is the interquartile range for Orlando's data set greater than the interquartile range for Victor's data set.

Step-by-step explanation:

Given:

Victor's Data Set: {22, 29, 32, 27, 30}.

Orlando's Data Set: {36, 25, 33, 27, 35}.

Now, to find the units is the interquartile range for Orlando's data set greater than the interquartile range for Victor's data set.

So, we get the interquartile range of Victor's Data Set:

{22,27,29,30,32}

Q3 = $\frac{30+32}{2}$

Q3 = $\frac{62}{2} =31.$

Q1 = $\frac{22+27}{2}$

Q1 = $\frac{49}{2}=24.5$

Thus, interquartile range is:

Interquartile range = $Q3-Q1$

Interquartile range = $31-24.5=6.5$

The interquartile range of Victor's Data Set = 6.5.

Now, to get the interquartile range of Orlando's Data Set:

{25,27,33,35,36}

Q1 = $\frac{25+27}{2}=\frac{52}{2} =26.$

Q3 = $\frac{35+36}{2} =\frac{71}{2}=35.5.$

Thus, interquartile range is:

Interquartile range = $Q3-Q1$

Interquartile range = $35.5-26=9.5$

The interquartile range of Orlando's Data set = 9.5.

Now, to get the units of the interquartile range for Orlando's data set greater than the interquartile range for Victor's data set we subtract Victor's interquartile range from Orlando's interquartile range:

$The\ interquartile\ range\ of\ Orlando's\ Data\ Set\ -\ The\ interquartile\ range\ of\ Victor's\ Data\ Set$

$=9.5-6.5=3.$

Therefore, by 3 units is the interquartile range for Orlando's data set greater than the interquartile range for Victor's data set.