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alejandra1201
31.07.2020 •
Mathematics
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.
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Ответ:
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,
= ![\frac{\sum X}{n}](/tpl/images/0715/9623/50881.png)
=
=
= 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.}](/tpl/images/0715/9623/5abec.png)
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}](/tpl/images/0715/9623/3783c.png)
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}](/tpl/images/0715/9623/3783c.png)
=![\frac{(\frac{8}{2})^{th} \text{ obs.} +(\frac{8}{2}+1)^{th} \text{ obs.} }{2}](/tpl/images/0715/9623/16e34.png)
=![\frac{(4)^{th} \text{ obs.} +(5)^{th} \text{ obs.} }{2}](/tpl/images/0715/9623/fa5b0.png)
=
= 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](/tpl/images/0715/9623/c09d4.png)
=![(\frac{8+1}{4} )^{th} \text{ obs.}](/tpl/images/0715/9623/e03a7.png)
=![(2.25 )^{th} \text{ obs.}](/tpl/images/0715/9623/168d5.png)
= 0 + 0.25[0 - 0] = 0
=![3(\frac{8+1}{4} )^{th} \text{ obs.}](/tpl/images/0715/9623/da318.png)
=![(6.75 )^{th} \text{ obs.}](/tpl/images/0715/9623/2a1f6.png)
= 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,
= ![\frac{\sum Y}{n}](/tpl/images/0715/9623/0229e.png)
=
=
= 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.}](/tpl/images/0715/9623/5abec.png)
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}](/tpl/images/0715/9623/3783c.png)
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}](/tpl/images/0715/9623/3783c.png)
=![\frac{(\frac{8}{2})^{th} \text{ obs.} +(\frac{8}{2}+1)^{th} \text{ obs.} }{2}](/tpl/images/0715/9623/16e34.png)
=![\frac{(4)^{th} \text{ obs.} +(5)^{th} \text{ obs.} }{2}](/tpl/images/0715/9623/fa5b0.png)
=
= 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](/tpl/images/0715/9623/c09d4.png)
=![(\frac{8+1}{4} )^{th} \text{ obs.}](/tpl/images/0715/9623/e03a7.png)
=![(2.25 )^{th} \text{ obs.}](/tpl/images/0715/9623/168d5.png)
= 1 + 0.25[2 - 1] = 1.25
=![3(\frac{8+1}{4} )^{th} \text{ obs.}](/tpl/images/0715/9623/da318.png)
=![(6.75 )^{th} \text{ obs.}](/tpl/images/0715/9623/2a1f6.png)
= 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.
Ответ:
An irrational number is a number which cannot be expressed in a ratio of two integers. In rational numbers, both numerator and denominator are whole numbers, where the denominator is not equal to zero. While an irrational number cannot be written in a fraction.S
Step-by-step explanation:
An irrational number is a number which cannot be expressed in a ratio of two integers. In rational numbers, both numerator and denominator are whole numbers, where the denominator is not equal to zero. While an irrational number cannot be written in a fraction.S