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gulaghasi [49]
3 years ago
6

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.
Mathematics
1 answer:
kkurt [141]3 years ago
5 0

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]

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