Chapter 10.2, Problem 3GP

i.

To determine

The fruit is range with the vegetable’s range.

Answer to Problem 3GP

The range of fruits (10) is smaller than that of vegetables (20).

Explanation of Solution

Given information: It shows the number of calories in a serving of certain fruits and vegetables.

Data set ₌$\{50,50,50,50,60,60,60,80\}$

Range = Highest-Lowest

  = $80-50=30$

Quartile $\{50,50,50,50\}$ , $\{60,60,60,80\}$

First Quartile = median $\{50,50,50,50\}=50$

Third quartile = Median $\{60,60,60,80\}=60$

Interquartile range = $60-50=10$

Outliers = $10\times 1.5=15$

Subtract 15 from the first quartiles and add 15 to the third quartiles $50-15=35,60+15=75$

Any value that is less than 35 or greater than 75 is an outlier $\{50,50,50,50,60,60,60,80\}$

Vegetables

Data set = $\{30,35,35,40,40,40,45,50\}$

Range = Highest-Lowest

  = $50-30=20$

Quartile $\{30,35,35,40\},\{40,40,45,50\}$

First quartile = 35

Third quartile = 42.5

Interquartile range = $42.5-35=7.5$

Outliers $7.5\times 1.5=11.25$

  $35-11.25=23.75,42.5+11.25=53.75$

  $\{30,35,35,40,40,40,45,50\}$

There are no outliers.

(a) The range of fruits (10) is smaller than that of vegetables (20).

ii.

To determine

To find: one of the outliers and outliers effect on the measure of variation for the numbers of calories in fruits.

Answer to Problem 3GP

The outlier increases both the median and mean.

Explanation of Solution

Given information: It shows the number of calories in a serving of certain fruits and vegetables.

80 is an outlier

Without the outlier

   Mean = $\frac{\text{sum}\text{of}\text{data}}{\text{7}}=54.3$

  Median = 50

With the outlier

   Mean = $\frac{\text{sum}\text{of}\text{data}+80}{8}=57.5$

  Median = 55

The outlier increases both the mean and the median.