Question

This data set represents the number of cups of coffee sold in a café between 8 a.m. and 10 a.m. every day for 14 days. {10, 9, 6, 12, 4, 6, 7, 8, 15, 14, 12, 9, 10, 5}. The difference of the values of the first and third quartiles of the data set is

Answer 1

first arrange in ascending order:-

4 5 6 6 7 8 9 9 10 10 12 12 14 15

first quartile = 6 and third quartile = 12

difference = 12-6 = 6

Answer 2

The difference of the values of the first and third quartiles of the data set is 6.

Explanation:

Data : 10, 9, 6, 12, 4, 6, 7, 8, 15, 14, 12, 9, 10, 5

This data set represents the number of cups of coffee sold in a café between 8 a.m. and 10 a.m. every day for 14 days.

Arrange the data in the ascending order :

4,5,6,6,7,8,9,9,10,10,12,12,14,15

No. of observations n = 14(even)

Median =
`\frac{((n)/(2)+1)^{\text{th term}}+(n)/(2)^{\text{th term}}}{2}`

So, Median =
`\frac{((14)/(2)+1)^{\text{th term}}+(14)/(2)^{\text{th term}}}{2}`

Median =
`\frac{8^{\text{th term}}+7^{\text{th term}}}{2}`

Median =
`(9+9)/(2)`

Median =
`(18)/(2)`

Median = 9

Now to find
`Q_3` i.e. Third quartile

Consider the set of values right to the median .

9,10,10,12,12,14,15

Now find the median of this data

No. of observations n = 7(odd)

So, Median =
`((n+1)/(2))^{\text{th term}}`

=
`((7+1)/(2))^{\text{th term}}`

=
`((8)/(2))^{\text{th term}}`

=
`(4)^{\text{th term}}`

=
`12`

So,
`Q_3=12`

Now to find
`Q_1` i.e. First quartile

Consider the set of values left to the median .

4,5,6,6,7,8,9

No. of observations n = 7(odd)

So, Median =
`((n+1)/(2))^{\text{th term}}`

=
`((7+1)/(2))^{\text{th term}}`

=
`((8)/(2))^{\text{th term}}`

=
`(4)^{\text{th term}}`

=
`6`

So,
`Q_1=6`

Now The difference of the values of the first and third quartiles of the data set : 12-6=6

Hence The difference of the values of the first and third quartiles of the data set is 6