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Statistics in Sports: Analyzing Olympic Athletes' Performance and Superstore Prize Draw - , Exámenes de Idioma Inglés

Solutions to statistical problems related to olympic athletes' sprinting times and a superstore prize draw. The first problem involves determining which athlete, peter or paul, has been more regular in their 100m sprinting times. The second problem deals with calculating the cumulative percentages, income percentages, and identifying the number of people who received prizes worth $20 less than expected in the superstore prize draw. Additionally, the document includes the calculation of the gini index and the determination of the number of people earning more than 1800 eur a month.

Tipo: Exámenes

2011/2012

Subido el 31/10/2012

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Introducci´on a la Estad´ıstica Curso 2012-2013 Fac. Comercio, Turismo y CC. Sociales Jovellanos
1. Two olympic athletes, Peter and Paul, recorded their 100m sprinting time in fifty rep-
etitions. Below, several statistics about the recorded times (in seconds) are provided:
arith. mean std. dev. percentile 25 percentile 50 percentile 75
Peter 10.30 0.06 10.26 10.31 10.34
Paul 10.22 0.10 10.15 10.22 10.29
Who has been more regular during the last 50 repetitions, Peter or Paul? Justify your
answer.
Solution.- Peter has been more regular than Paul (Peter’s sprint times have been less
spread out than those of Paul). We can note it by looking of the values of all the available
dispersion measures. The values of the standard deviation, coefficient of variation,
interquartile range and variance are lower in Peter’s records than in Paul’s ones. Those
values are provided below:
std. dev. I.R. C.V. Variance
Peter 0.06s 10.34s-10.26s=0.08s 0.06
10.30 0.00583 0.036s2
Paul 0.10s 10.29s-10.15s=0.14s 0.10
10.22 0.00978 0.01s2
2. A superstore has organized a prize draw. Prizes range from $5 to $100 and they are
distributed as follows:
10 white gift cards, $5 each.
5 red gift cards, $20 each.
3 silver gift cards, $50 each.
2 golden gift cards, $100 each.
(a) Complete the following table:
xiniNiPi(cum %) xiniPi
j=1 xjnjQi(cum. % of income)
5 10 10 50% 50 50 10%
20 5
50 3
100 2
Solution.- The complete table is:
xiniNiPi(cum %) xiniPi
j=1 xjnjQi(cum. % of income)
5 10 10 50% 50 50 10%
20 5 15 75% 100 150 30%
50 3 18 90% 150 300 60%
100 2 20 100% 200 500 100%
(b) How much money has been given out? (What has been the total amount of money
spent by the superstore?) The total amount of money is P4
j=1 xini= $500.
pf3
pf4

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Introducci´on a la Estad´ıstica Curso 2012-2013 Fac. Comercio, Turismo y CC. Sociales Jovellanos

  1. Two olympic athletes, Peter and Paul, recorded their 100m sprinting time in fifty rep- etitions. Below, several statistics about the recorded times (in seconds) are provided:

arith. mean std. dev. percentile 25 percentile 50 percentile 75 Peter 10.30 0.06 10.26 10.31 10. Paul 10.22 0.10 10.15 10.22 10.

Who has been more regular during the last 50 repetitions, Peter or Paul? Justify your answer. Solution.- Peter has been more regular than Paul (Peter’s sprint times have been less spread out than those of Paul). We can note it by looking of the values of all the available dispersion measures. The values of the standard deviation, coefficient of variation, interquartile range and variance are lower in Peter’s records than in Paul’s ones. Those values are provided below:

std. dev. I.R. C.V. Variance Peter 0.06s 10.34s-10.26s=0.08s 100.^06. 30 ≈ 0. 00583 0. 036 s^2 Paul 0.10s 10.29s-10.15s=0.14s 100.^10. 22 ≈ 0. 00978 0. 01 s^2

  1. A superstore has organized a prize draw. Prizes range from $5 to $100 and they are distributed as follows: - 10 white gift cards, $5 each. - 5 red gift cards, $20 each. - 3 silver gift cards, $50 each. - 2 golden gift cards, $100 each.

(a) Complete the following table:

xi ni Ni Pi (cum %) xini

∑i j=1 xj^ nj^ Qi^ (cum. % of income) 5 10 10 50% 50 50 10% 20 5 50 3 100 2

Solution.- The complete table is:

xi ni Ni Pi (cum %) xini

∑i j=1 xj^ nj^ Qi^ (cum. % of income) 5 10 10 50% 50 50 10% 20 5 15 75% 100 150 30% 50 3 18 90% 150 300 60% 100 2 20 100% 200 500 100% (b) How much money has been given out? (What has been the total amount of money spent by the superstore?) The total amount of money is

j=1 xini^ = $500.

(c) How many people have received a prize valued at $20 less? Solution.- They are 15 people (10+5=15). What percentage of the awarded people do they represent? Solution.- The total number of awarded people is N 4 = n = 20. Thus, the above 15 people represent 75% of them. What percentage of the total amount of money did they (altogether) receive? Solution.- They (altogether) received $150. (Cumulative income associated to the second value in the sample.) This amount of money represents Q 2 = 30% of the total. (d) Select the appropriate value for the Gini index from the following list. Justify your choice: IG = 0, IG = 0. 05 , IG = 1, IG = − 0. 60 , IG = 0. 53 Solution.- Gini index is always between 0 and 1, and therefore the third option is not possible. Furthermore, in this example, it is not null because there is not total equality. (There are different values of prizes, ranging from $5 to $100), either it is equal to 1 (There is not total equality.) Furthermore, the value 0. 05 does not seem to be in accordance with the above data, because the level of inequality seems to be higher than that. The only possibility is therefore IG = 0. 53. We can check it, anyway, by calculating the index, according to the following formula:

IG =

i=1 ∑(Pi^ −^ Qi) 3 i=1 Pi

  1. The following joint frequency table provides information about the monthly salary and savings in a sample of workers:

X =salary\Y =saving 0 100 200 500 900 900 3 2 0 0 0 1500 1 3 2 0 0 2000 1 3 3 1 0 3000 0 1 2 2 1

(a) How many people are there in the sample? Solution.- There are 25 people in the sample (the sum of all joint frequencies.) (b) How many people in the sample earn more than 1800 Eur a month? Solution.- The number of people that earn more than 1800 Eur is the number of people that earn 2000 or 3000 Eur a month, 1 + 3 + 3 + 1 + 0 + 0 + 1 + 2 + 2 + 1 = 14. (c) Out of those who earn more than 1800 Eur a month: i. What is the arithmetic mean of their savings? Solution.- Let us first determine the frequency table of their savings: yj n′ .j 0 1 100 4 200 5 500 3 900 1

  • The third scatter plot corresponds to a pair of null covariance and null coefficient of determination: there is not an increasing, either a decreasing relation between both variables.