Question

Item 4 Find the median, first quartile, third quartile, and interquartile range of the data. 40,33,37,54,41,34,27,39,35 Item 4 Find the median, first quartile, third quartile, and interquartile range of the data. 40,33,37,54,41,34,27,39,35

Answer 1

Median =37

First quartile = 33.5

Third quartile=40.5

Interquartile=7

Explanation:

Given data

40,33,37,54,41,34,27,39,35

First we have to rearrange the data in either ascending order or descending order.

We can rewrite the data as

27,33,34,35,37,39,40,41,54

Median:

The middle term of the data is the median of the data.

The number of data = n

If n is odd, the median of the data=
`(\frac{ n+1}2)^{th` term

If n is even, the median of the data
`\frac{(\frac {n}{2})^(th) +{(\frac {n}{2}+1)^(th)}}{2}`

Here n= 9

The median of the data=
`((9+1)/(2))^(th)` term

`=((10)/(2))^(th)` term

`=5^(th)` term

=37

First quartile:

The median term of the lower half of the data.

Lower half = 27,33,34,35

The median of the lower half =
`((\frac 42)^(th) term+(\frac 42+1)^(th)term)/(2)`

=
`((2)^(nd) term+(3)^(rd)term)/(2)`

=
`\frac{33+34}2`

=33.5

The first quartile= 33.5

Third quartile:

The median term of the upper half of the data.

Upper half 39,40,41,54

The median of the lower half =
`((\frac 42)^(th) term+(\frac 42+1)^(th)term)/(2)`

=
`((2)^(nd) term+(3)^(rd)term)/(2)`

=
`\frac{40+41}2`

=40.5

Interquartile:

The difference between first quartile and third quartile.

Interquartile = 40.5-33.5

=7