Chapter 12.2, Problem 30HP

a.

To determine

To find a set of numbers whose mean, median and mode are equal.

Answer to Problem 30HP

The set of a numbers whose mean, median and mode are equal is $\left\{2,5,5,6,7\right\}$ .

Explanation of Solution

Given information :

The numbers whose mean, median and mode are equal.

Formula used :

Mean of a set of number, $x_1,x_2,x_3,\mathrm{........}{x}_n$ is defined by,

  $\frac{x_1+x_2+x_3+\mathrm{.......}+x_n}{n}$

The Median is:

  $x_{\left(\raisebox{1ex}{$n+1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}$ [if n is odd]

  $\frac{x_{\left(\raisebox{1ex}{$n$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}+x_{\left(\raisebox{1ex}{$n$}\!\left/ \!\raisebox{-1ex}{$2$}\right.+1\right)}}{2}$ [if n is even]

The mode is the number in the list that occurs most often.

Calculation :

Assume the set of the number is, $\left\{2,5,5,6,7\right\}$ .

Mean of the number is obtained as:

  $\frac{2+5+5+6+7}{5}=\frac{25}{5}=5$

Median is obtained as:

  $x_{\left(\raisebox{1ex}{$5+1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}=x_3=5$ [since, $n=5$ which is odd]

Mode is 5 since it occurs more often.

Hence,

The set of a numbers whose mean, median and mode are equal is $\left\{2,5,5,6,7\right\}$

b.

To determine

To find a set of numbers whose mean is greater than the median.

Answer to Problem 30HP

The set of a numbers whose mean is greater than the median is $\left\{2,3,4,7,9\right\}$ .

Explanation of Solution

Given information :

The numbers whose mean is greater than the median.

Formula used :

Mean of a set of number, $x_1,x_2,x_3,\mathrm{........}{x}_n$ is defined by,

  $\frac{x_1+x_2+x_3+\mathrm{.......}+x_n}{n}$

The Median is:

  $x_{\left(\raisebox{1ex}{$n+1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}$ [if n is odd]

  $\frac{x_{\left(\raisebox{1ex}{$n$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}+x_{\left(\raisebox{1ex}{$n$}\!\left/ \!\raisebox{-1ex}{$2$}\right.+1\right)}}{2}$ [if n is even]

Calculation :

Assume the set of the number is, $\left\{2,3,4,7,9\right\}$ .

Mean of the number is obtained as:

  $\frac{2+3+4+7+9}{5}=\frac{25}{5}=5$

Median is obtained as:

  $x_{\left(\raisebox{1ex}{$5+1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}=x_3=4$ [since, $n=5$ which is odd]

So, mean is greater than median.

Hence,

The set of a numbers whose mean is greater than the median is $\left\{2,3,4,7,9\right\}$

c.

To determine

To find a set of numbers whose mode is 10 and the median is greater than the mean

Answer to Problem 30HP

The set of a numbers whose mode is 10 and the median is greater than the mean is $\left\{1,5,10,10,14\right\}$ .

Explanation of Solution

Given information :

The numbers whose mode is 10 and the median is greater than the mean.

Formula used :

Mean of a set of number, $x_1,x_2,x_3,\mathrm{........}{x}_n$ is defined by,

  $\frac{x_1+x_2+x_3+\mathrm{.......}+x_n}{n}$

The Median is:

  $x_{\left(\raisebox{1ex}{$n+1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}$ [if n is odd]

  $\frac{x_{\left(\raisebox{1ex}{$n$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}+x_{\left(\raisebox{1ex}{$n$}\!\left/ \!\raisebox{-1ex}{$2$}\right.+1\right)}}{2}$ [if n is even]

The mode is the number in the list that occurs most often.

Calculation :

Assume the set of the number is, $\left\{1,5,10,10,14\right\}$ .

Mean of the number is obtained as:

  $\frac{1+5+10+10+14}{5}=\frac{40}{5}=8$

Median is obtained as:

  $x_{\left(\raisebox{1ex}{$5+1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}=x_3=10$ [since, $n=5$ which is odd]

So, median is greater than mean.

Mode is 10 since it occurs more often.

Hence,

The set of a numbers whose mode is 10 and the median is greater than the mean is $\left\{1,5,10,10,14\right\}$

d.

To determine

To find a set of numbers whose mean is 6, median is 5.5 and mode is 9

Answer to Problem 30HP

The set of a numbers whose mean is 6, median is 5.5 and mode is 9 is $\left\{3,4,5,6,9,9\right\}$

Explanation of Solution

Given information :

The numbers whose mean is 6, median is 5.5 and mode is 9

Formula used :

Mean of a set of number, $x_1,x_2,x_3,\mathrm{........}{x}_n$ is defined by,

  $\frac{x_1+x_2+x_3+\mathrm{.......}+x_n}{n}$

The Median is:

  $x_{\left(\raisebox{1ex}{$n+1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}$ [if n is odd]

  $\frac{x_{\left(\raisebox{1ex}{$n$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}+x_{\left(\raisebox{1ex}{$n$}\!\left/ \!\raisebox{-1ex}{$2$}\right.+1\right)}}{2}$ [if n is even]

The mode is the number in the list that occurs most often.

Calculation :

Assume the set of the number is, $\left\{3,4,5,6,9,9\right\}$ .

Mean of the number is obtained as:

  $\frac{3+4+5+6+9+9}{6}=\frac{36}{5}=6$

Median is obtained as:

  $\frac{x_{\left(\raisebox{1ex}{$6$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}+x_{\left(\raisebox{1ex}{$6$}\!\left/ \!\raisebox{-1ex}{$2$}\right.+1\right)}}{2}=\frac{x_3+x_4}{2}=\frac{5+6}{2}=5.5$ [since, $n=6$ which is even]

So, median is greater than mean.

Mode is 9 since it occurs more often.

Hence,

The set of a numbers whose mean is 6, median is 5.5 and mode is 9 is $\left\{3,4,5,6,9,9\right\}$ .