F.3 Measures of Central Tenden

[匿名]

2015-01-09 6:56 am

There are 40 students in F3A.In a mathematics test,the mean score of the whole
class is 52.4.The mean score for boys is 8 less than that of girls.Let X be the number of boys and Y be the mean score of boys.

a) Show that 5Y-X=222
b) If P is the sum of both mean scores of the boys and girls, find the minimum value
of P.

回答 (1)

Ben

2015-01-09 9:31 pm

✔ 最佳答案

The number of boys is X, then the number of girls is (40-X)
The mean score of boys Y, then the mean score of girls is (Y+8)
(a)
(X*Y+(40-X)*(Y+8))/40=52.4
XY+40Y+320-XY-8X=2096
40Y-8X=1776
8(5Y-X)=1776
5Y-X=222
(b)
P=Y+(Y+8)
P=2Y+8
From(a),Y=(222+X)/5
P=2((222+X)/5)+40/5
P=(444+2X+40)/5
P=(484+2X)/5
In F3A, there is at least one boy
So, the minimum number of boy is 1, then X=1
P=(484+2(1))/5
P=486/5
P=97.2
So, the minimum value of P is 97.2

2015-01-09 13:34:47 補充：
In(b),it is impossible that X=0
If X=0, then Y must be 0.
Put X=0 and Y=0 into 5Y-X=222 is impossible.
So the value of X must be greater than 0.

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