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

Answer:

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

Step-by-step 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, $\bar X$ =  $\frac{\sum X}{n}$

                       =  $\frac{11+ 3+ 0+ 0+ 2+ 0+ 6+ 4}{8}$  =  $\frac{26}{8}$  = 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  =  $(\frac{n+1}{2} )^{th} \text{ obs.}$

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

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

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

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

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

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

                   =  $\frac{2+3}{2}$  = 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 = $Q_3-Q_1$

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

     =  $(\frac{8+1}{4} )^{th} \text{ obs.}$

     =  $(2.25 )^{th} \text{ obs.}$

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

     =  0 + 0.25[0 - 0] = 0

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

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

     =  $(6.75 )^{th} \text{ obs.}$

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

     =  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, $\bar Y$ =  $\frac{\sum Y}{n}$

                       =  $\frac{4+ 2+ 0+ 3+ 2+ 1+ 6+ 4}{8}$  =  $\frac{22}{8}$  = 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  =  $(\frac{n+1}{2} )^{th} \text{ obs.}$

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

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

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

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

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

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

                   =  $\frac{2+3}{2}$  = 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 = $Q_3-Q_1$

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

     =  $(\frac{8+1}{4} )^{th} \text{ obs.}$

     =  $(2.25 )^{th} \text{ obs.}$

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

     =  1 + 0.25[2 - 1] = 1.25

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

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

     =  $(6.75 )^{th} \text{ obs.}$

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

     =  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.