What is the probability that out of 125 babies born, at least 60 will be girls? Assume that boys and girls are equally probable. (I got the answer 70.54 I just want to make sure I did it correctly, thanks!)
we can use the formula we used in the previous problem
we want probability 60 or greater
ok thanks, what i did this time was i did 125(.5)= 62.5 sqrt(31.25) then: z=60-62.5/sqrt(31.25)= -.4472135955 and then looked it up on the z score chart and thats how i got the answer
right, just make sure you know what type of z table it is sometimes it uses P ( Z < a ) , sometimes it uses P ( Z > a) if it is the latter, you will need to subtract by 1
ohh ok so is that where i messed up?
also this is where i got the table from http://www.stat.ufl.edu/~athienit/Tables/Ztable.pdf
yes that table gives you P ( Z < a )
also you need to make a binomial continuity correction
we are going to find P( X >= 59.5 ) , since X = 60 is a rectangle from 59.5 to 60.5
ok cool so do i just replace the 60 with 59.5 in the same equation?
yes, to find the z score
it turns out on calculator normalcdf ( 60, 1E99, 125*.5 , sqrt( 125 * .5 * .5) ) = .6726 , which is wrong
thats why we need the 'binomial correction factor '
yea i replaced it at 59.5-62.5/5.59 and got 3/5.59= .5366726
that should be -.5367
and now P( Z > -.5367) = 1 - P ( Z < -.5367) by complement rule
ok, so i shouldn't use the z-score table but ue the complement rule instead?
youre doing it correct
we just have to use complement rule, since the table only shows one direction
lets round -.5367 to -.54 P( Z < -.54) = .2981 now we want P( Z > -.54) but that is equal to 1 - P ( Z < -.54) = 1- .2946 this is only accurate to 2 decimal places i would say
ok so it is 70.54 then
isn't statistics fun :)
let me draw what we are doing
hahaha, that was the answer i got in the beginning!
yes
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