Question

The number of pets 40 students own is recorded

Number of pets: 2 , 4, 6, 8, 10
Number of students: x, 2, y, 6, 14

a) (i) show that x+y=18
(ii) If the mean of the distribution is 6.4, show that x +3y =30
(iii) Hence, find the value of x and of y
b) Using your answers in (a) (iii), find
(i) the median,
(ii) the mode,
Of the distribution

Answer 1

Answer:

a.

i) & ii) -> See Explanation Below

a.

iii) $y = 6$ and $x = 12$

b.

i) $Median = 7$

ii) $Mode = 10$

Step-by-step explanation:

Given

Number of pets: 2 , 4, 6, 8, 10

Number of students: x, 2, y, 6, 14

Total students = 40

Required

a) (i) show that x+y=18

Given that total number of students is 40;

This implies that

$x + 2 + y + 6 + 14 = 40$

Collect like terms

$x + y = 40 - 14 - 6 - 2$

$x + y = 18$

a) (ii) If the mean of the distribution is 6.4, show that x +3y =30

The mean of a distribution is calculated as thus

$Mean = \frac{\sum fx}{\sum x}$

$\sum fx}$ is gotten by multiplying number of pets by corresponding number of students

$\sum fx} = 2 * x + 4 * 2 + 6 * y + 8 * 6 + 10 * 14$

$\sum fx} = 2 x + 8 + 6y + 48 + 140$

$\sum fx} = 2 x + 6y + 48 + 140+ 8$

$\sum fx} = 2 x + 6y + 196$

$\sum x}$ is the total number of students

$\sum x} = 40$

So;

$Mean = \frac{\sum fx}{\sum x}$ becomes

$Mean = \frac{2 x + 6y + 196}{40}$

Substitute 6.4 for Mean

$6.4 = \frac{2 x + 6y + 196}{40}$

Multiply both sides by 40

$256 = 2 x + 6y + 196$

Subtract 196 from both sides

$256 -196= 2 x + 6y + 196-196$

$60= 2 x + 6y$

Divide both sides by 2

$\frac{60}{2}= \frac{2 x}{2} + \frac{6y}{2}$

$30 = \frac{2 x}{2} + \frac{6y}{2}$

$30 = x + 3y$

Reorder

$x + 3y = 30$

a) (iii) Hence, find the value of x and of y

Using

$x + y = 18$ and

$x + 3y = 30$

Subtract both equations

$(x + y = 18) - (x + 3y = 30)$

$x - x + y - 3y = 18 - 30$

$y - 3y = 18 - 30$

$-2y = -12$

Divide both sides by -2

$\frac{-2y}{-2} = \frac{-12}{-2}$

$y = \frac{-12}{-2}$

$y = 6$

Substitute 6 for y in $x + y = 18$

$x + y = 18$ becomes

$x + 6 = 18$

Subtract 6 from both sides

$x + 6 - 6 = 18 - 6$

$x = 12$

b.

i) Calculate the Median

First, we need to tabulate the given data properly

Number of Pets ------ Number of students ------- Cumulative Frequency

2 ------------------------------12 -----------------------------------12

4 ------------------------------2 -----------------------------------14

6 ------------------------------6 -----------------------------------20

8 ------------------------------6 -----------------------------------26

10 ------------------------------14 -----------------------------------40

Since the number of pets is already arranged in ascending order,

the next step is to calculate the median element

Number of students = 40

$Median = \frac{40}{2}$

Median = 20

Given that number of students (40) is an even number,

The median is the average of the 20th and 21st element

From the table above; the median can be gotten from

6 ------------------------------6 -----------------------------------20

8 ------------------------------6 -----------------------------------26

The 20th element = 6

The 21st element = 8

$Median = \frac{6 + 8}{2}$

$Median = \frac{14}{2}$

$Median = 7$

ii) Mode

The mode is the corresponding data with the highest frequency

The highest frequency is 14;

The number of pets with frequency of 14 is 10

Hence,

$Mode = 10$