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Q)Two university female basketball players are being considered for an award to be given to the most consistent player with the highest scoring average for the season. The number of points scored per game for all games played during the season is given by the following table. ( Table Attached)
a) Calculate the population mean and population standard deviation of the number of points scored per game for each player.
(b) Using the results of part (a) which player should receive the award.
Answer:
<h3><u>Player B should receive the award.</u></h3><h3>Step-by-step explanation:</h3>Points of player A:- 5,32,45,25,8,10,35,12,37,21
<u>First we will calculate the mean for player A</u>
∑x = 5 + 32 + 45 + 25 + 8 + 10 + 35 + 12 + 37 + 21
∑x = 230
N = 10
<u>Mean = µ= ∑x/N = 230/10 = 23</u>
<u>Now we will calculate the standard deviation for Player A</u>
x - µ = -18 , 9, 22, 2, -15, -13, 12, 11, 14, -2
(x - µ)^2 = 324, 81, 484, 4, 225, 169, 144, 121, 196, 4
∑(x - µ)^2 = 1752
σ = sqrt( ∑((x - µ)^2/N) )
σ = sqrt(1752/10)
<u>σ = 13.236</u>
<u>Calculating mean for player B:</u>
Points of player B = 18, 20, 22, 15, 35, 24, 29, 19, 25, 23
∑x = 18 + 20 + 22 + 15 + 35 + 24 + 29 + 19 + 25 + 23
N =10
µ = 230/10
<u>µ = 23</u>
<u>Calculating standard deviation for player B:</u>
(x - µ ) = -5, -3, -1, -8, 12, 1, 6, -4, 2, 0
(x -µ )^2 = 25, 9, 1, 64, 144, 1, 36, 16, 4, 0
∑(x - µ)^2 = 300
σ = sqrt(300/10)
<u>σ = 5.48</u>
Since both Player A and Player B have same mean we will calculated the coefficient of variation for both the players.
<u>For player A:</u>
σa/µa *100 = 13.24/23 *100 = 57.56
<u>For Player B:</u>
σb/µb *100 = 5.48/23 * 100 = 23.82
<u>Since the coefficient of variation of Player B is less than Player A hence Player B should receive the award.</u>
