prob(win) = .1
prob(not win) = .9
prob(at least 5 of 50)
= 1 - (prob(none of 50) + prob(1 of 50) + prob(2 of 50) + prob(3of50) + prob(4of50) )
= 1 - (.9^50 + 50(.1) (.9)^49 + 1225(.1)^2 (.9)^48 + 19600(.1)^3 (.9)^47 + 230300(.1)^4 (.9)^46 )
= .5688
check my arithmetic, I did it twice
The probability of winning a large stuffed animal in the ring-toss game at a fair is 10%. Find the probability of winning at least 5 of 50 games using normal approximation. I got 0.4052 but answers says it's 0.5932.
Please help!
4 answers
mean = 5
sd = sqrt(5*.9) = 2.121
look for greater than 4.5 !!!!!
use eg
http://davidmlane.com/hyperstat/z_table.html
sd = sqrt(5*.9) = 2.121
look for greater than 4.5 !!!!!
use eg
http://davidmlane.com/hyperstat/z_table.html
So you have to do 1 minus 0.4052 to get your answer?
Most tables give the area below the normal distribution curve on the left (<Z). In that case, yes, you need to subtract.
The link Damon gave you has the option of finding area to the right, (>Z). In that case, you do not have to subtract from 1.
Also .4052 is straight reading of Z=-0.24, rounded from -0.5/sqrt(4.5)=-.5/2.12132=0.2357.
Even with paper tables, you could refine the reading by linear interpolation .4090*.43+.4052*.57=.4068, which will give 1-.4068=.5932, the exact book answer.
The link Damon gave you has the option of finding area to the right, (>Z). In that case, you do not have to subtract from 1.
Also .4052 is straight reading of Z=-0.24, rounded from -0.5/sqrt(4.5)=-.5/2.12132=0.2357.
Even with paper tables, you could refine the reading by linear interpolation .4090*.43+.4052*.57=.4068, which will give 1-.4068=.5932, the exact book answer.