Question

Two soccer teams play 8 games in their season. The number of goals each team scored per game is listed below: Team X: 11, 3, 0, 0, 2, 0, 6, 4 Team Y: 4, 2, 0, 3, 2, 1, 6, 4 Which of the following is true? A. Team X’s scores have a lower interquartile range. B. Team X’s scores have a higher median value. C. Team Y’s scores have a lower mean value. D. Both teams have the same range of scores.

Answer 1

C. Team Y’s scores have a lower mean value.

Explanation:

We are given that Two soccer teams play 8 games in their season. The number of goals each team scored per game is listed below:

Team X: 11, 3, 0, 0, 2, 0, 6, 4

Team Y: 4, 2, 0, 3, 2, 1, 6, 4

Firstly, we will calculate the mean, median, range and inter-quartile range for Team X;

Mean of Team X data is given by the following formula;

Mean,
`\bar X` =
`(\sum X)/(n)`

=
`(11+ 3+ 0+ 0+ 2+ 0+ 6+ 4)/(8)` =
`(26)/(8)` = 3.25

So, the mean of Team X's scores is 3.25.

Now, for calculating the median; we have to arrange the data in ascending order and then observe that the number of observations (n) in the data is even or odd.

Team X: 0, 0, 0, 2, 3, 4, 6, 11

• If n is odd, then the formula for calculating median is given by;

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

• If n is even, then the formula for calculating median is given by;

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

Here, the number of observations is even, i.e. n = 8.

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

=
`\frac{((8)/(2))^(th) \text{ obs.} +((8)/(2)+1)^(th) \text{ obs.} }{2}`

=
`\frac{(4)^(th) \text{ obs.} +(5)^(th) \text{ obs.} }{2}`

=
`(2+3)/(2)` = 2.5

So, the median of Team X's score is 2.5.

Now, the range is calculated as the difference between the highest and the lowest value in our data.

Range = Highest value - Lowest value

= 11 - 0 = 11

So, the range of Team X's score is 11.

Now, the inter-quartile range of the data is given by;

Inter-quartile range =
`Q_3-Q_1`

`Q_1=((n+1)/(4) )^(th) \text{ obs.}`

=
`((8+1)/(4) )^(th) \text{ obs.}`

=
`(2.25 )^(th) \text{ obs.}`

`Q_1 = 2^(nd) \text{ obs.} + 0.25[ 3^(rd) \text{ obs.} -2^(nd) \text{ obs.} ]`

= 0 + 0.25[0 - 0] = 0

`Q_3=3((n+1)/(4) )^(th) \text{ obs.}`

=
`3((8+1)/(4) )^(th) \text{ obs.}`

=
`(6.75 )^(th) \text{ obs.}`

`Q_3 = 6^(th) \text{ obs.} + 0.75[ 7^(th) \text{ obs.} -6^(th) \text{ obs.} ]`

= 4 + 0.75[6 - 4] = 5.5

So, the inter-quartile range of Team X's score is (5.5 - 0) = 5.5.

Now, we will calculate the mean, median, range and inter-quartile range for Team Y;

Mean of Team Y data is given by the following formula;

Mean,
`\bar Y` =
`(\sum Y)/(n)`

=
`(4+ 2+ 0+ 3+ 2+ 1+ 6+ 4)/(8)` =
`(22)/(8)` = 2.75

So, the mean of Team Y's scores is 2.75.

Now, for calculating the median; we have to arrange the data in ascending order and then observe that the number of observations (n) in the data is even or odd.

Team Y: 0, 1, 2, 2, 3, 4, 4, 6

• If n is odd, then the formula for calculating median is given by;

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

• If n is even, then the formula for calculating median is given by;

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

Here, the number of observations is even, i.e. n = 8.

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

=
`\frac{((8)/(2))^(th) \text{ obs.} +((8)/(2)+1)^(th) \text{ obs.} }{2}`

=
`\frac{(4)^(th) \text{ obs.} +(5)^(th) \text{ obs.} }{2}`

=
`(2+3)/(2)` = 2.5

So, the median of Team Y's score is 2.5.

Now, the range is calculated as the difference between the highest and the lowest value in our data.

Range = Highest value - Lowest value

= 6 - 0 = 6

So, the range of Team Y's score is 6.

Now, the inter-quartile range of the data is given by;

Inter-quartile range =
`Q_3-Q_1`

`Q_1=((n+1)/(4) )^(th) \text{ obs.}`

=
`((8+1)/(4) )^(th) \text{ obs.}`

=
`(2.25 )^(th) \text{ obs.}`

`Q_1 = 2^(nd) \text{ obs.} + 0.25[ 3^(rd) \text{ obs.} -2^(nd) \text{ obs.} ]`

= 1 + 0.25[2 - 1] = 1.25

`Q_3=3((n+1)/(4) )^(th) \text{ obs.}`

=
`3((8+1)/(4) )^(th) \text{ obs.}`

=
`(6.75 )^(th) \text{ obs.}`

`Q_3 = 6^(th) \text{ obs.} + 0.75[ 7^(th) \text{ obs.} -6^(th) \text{ obs.} ]`

= 4 + 0.75[4 - 4] = 4

So, the inter-quartile range of Team Y's score is (4 - 1.25) = 2.75.

Hence, the correct statement is:

C. Team Y’s scores have a lower mean value.