Edexcel S2: Binomial-Normal Approximations Question
I just don't see what's happening in the following:
The Random Variable X~B(100, 0.65). Use a suitable approximation to estimate P(60<X<72).
My Working:
Mean=100*0.65=65, S.D.=sqrt(100*0.65*0.35)=sqrt(91)/2
X~N(65, 22.75)
P(60<X<72)≈P(59.5<Y<72.5)
P[(59.5-65/sqrt(91)/2)<Z<(72.5-65/sqrt(91)/2)]
P(-1.15<Z<1.57)
Referring to the normal distribution table
0.9418-(1-0.8749)=0.8167
The markscheme says 0.768~0.769. I think this means any value between these values. Help!!!!!!!!
The Random Variable X~B(100, 0.65). Use a suitable approximation to estimate P(60<X<72).
My Working:
Mean=100*0.65=65, S.D.=sqrt(100*0.65*0.35)=sqrt(91)/2
X~N(65, 22.75)
P(60<X<72)≈P(59.5<Y<72.5)
P[(59.5-65/sqrt(91)/2)<Z<(72.5-65/sqrt(91)/2)]
P(-1.15<Z<1.57)
Referring to the normal distribution table
0.9418-(1-0.8749)=0.8167
The markscheme says 0.768~0.769. I think this means any value between these values. Help!!!!!!!!
Original post by Nuclear Ghost
P(60<X<72)≈P(59.5<Y<72.5)
P(60<X<72)≈P(59.5<Y<72.5)
This bit is in error. Since X > 60, then your lower limit should be 60.5
Original post by ghostwalker
This bit is in error. Since X > 60, then your lower limit should be 60.5
Thanks again, mate.
Heyyy your contiunity correction is wrong, it is P(60.5_<Y< 72.5). try using these and your answer will be 0.7682 like the markscheme 
Original post by Dhruvi_majithia
Heyyy your contiunity correction is wrong, it is P(60.5_<Y< 72.5). try using these and your answer will be 0.7682 like the markscheme
I suspect after 6 years the OP wouldn't care even if someone else hadn't pointed it out at the time...
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