Question

Here is some information about 26 houses.a,b and c are all different numbers.Number of bedrooms:1,2,3,4,5.Number of houses:7,a,b,c,8.The median number of bedrooms is 3.5 Work out a possible set of values for a,b and c.

Answer 1

Answer:

$[a,b,c]=[4,2,5]$ or $[a,b,c]=[2,4,5]$

Step-by-step explanation:

Given

$\begin{array}{cccccc}{Bedroom} & {1} & {2} & {3} & {4} & {5} \ \\ {Houses} & {7} & {a} & {b} & {c} & {8} \ \end{array}$

$n = 26$

$Median = 3.5$

Required

Find a, b and c

Median is calculated as:

$Median = \frac{n+1}{2}$

$Median = \frac{26+1}{2}$

$Median = \frac{27}{2}$

$Median = 13.5th$

This means that the median is the average of the 13th and 14th item.

$\begin{array}{cccccc}{Bedroom} & {1} & {2} & {3} & {4} & {5} \ \\ {Houses} & {7} & {a} & {b} & {c} & {8} \ \end{array}$

Since the median is $Median = 3.5$ (average of 3 and 4),

and:

$3 \to b \to 13$

$4 \to c \to 14$

This implies that c begins at 14

So, the number of houses bedrooms 4 and 5 is:

$c + 8 = 26 - 14 +1$

$c + 8 = 13$

$c = 5$

Also

$7 + a + b + c + 8 = 26$ ---- sum of total frequency

$a + b + c = 26 - 7 - 8$

$a + b + c = 11$

Substitute $c = 5$

$a + b + 5 = 11$

$a + b = 11 - 5$

$a + b = 6$

Since a, b and c are different, then

a and b cannot be 5 because $c = 5$

a and b cannot be 3 because $a = b =3$

$b \ne 0$ because b ends at 13

a

So, possible sets are:

$[a,b]=[2,4]$

$[a,b]=[4,2]$

Include c, we have:

$[a,b,c]=[4,2,5]$ or $[a,b,c]=[2,4,5]$