∴x=6,y=7；
(2) The 2× 2 contingency table is as follows:
School A and School B total excellent 10 20 30 non-excellent 45 30 75 total 55 50105 ÷ k2 =105× (10× 30? 20×45)230×75×50×55≈6. 109>5.024,
∴ 97.5% confirmed that there were differences in mathematics scores between the two schools;
(3) The excellent rate of A school is 2 1 1, and that of B school is 25.
ξ=0， 1，2，3，ξ~B(3，25)
P(ξ=0)=C03(25)0( 1？ 25)3=27 125; p(ξ= 1)= c 13(25) 1( 1？ 25)2=54 125;
P(ξ=2)=C23(25)2( 1？ 25) 1=36 125; P(ξ=3)=C33(25)3( 1？ 25)0=8 125,
Distribution list of ξ is
? ξ ? 0 ? 1 ? 2 ? 3? p？ 27 125 ? 54 125 ? 36 125 ? I praised it and stepped on it. What is your evaluation of this answer? Comments put away Shenzhen Fields Education
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