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.

or

Step-by-step explanation:

Given

Required

Find a, b and c

Median is calculated as:

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

Since the median is (average of 3 and 4),

and:

This implies that c begins at 14

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

Also

---- sum of total frequency

Substitute

Since a, b and c are different, then

a and b cannot be 5 because

a and b cannot be 3 because

because b ends at 13

a

So, possible sets are:

Include c, we have:

or

c

step-by-step explanation:

it isn't d, because the interquartile range is from 27-57. and the interquartile range is basically half the data, it's pretty obvious.

it isn't b, because like i said earlier, the interquartile range is from 27-57. 57-27 is 30. the real range has to be longer

not a either. although it looks like it, 3/4 of the data is not greater than 27. but if c is wrong than this should be correct

the correct answer is c, because 13 is the point in which the plot starts

i found two solutions,   n = 2 • ± √3 = ± 3.4641

step-by-step explanation: