Help the problem is due in the morning.
I did post this in the urgent math help section, but I did not get an answer, so I am trying here.
I am going to try and delete my post in the Urgent section so I don't have duplicates.
Thanks all you math geniuses!
Use the binomial distriubution with n=20, p = 0.5 to compare accuracy of the normal approximation to the binomial.
a. Compute the exact probablities and corresponding normal approximations for y<2.
b. The normal approximation can be improved slightly by taking P(y<2.5).
Compare your results. Why should this help?
c. Compute the exact probabilities and corresponding normal approximations for 3<y<6 (note: 3.5<y<6.5 seems to work better for the normal approximation.)
Note: (0.5) to the 20th = 0.000001)
I did the first part calculating the exact propabilities using the binomial probability disribution.
P(y<2) = P(y=1) + P(y=0)
P(1) = [20!/1!(20-1)!] (0.5)^1 (0.5)^19 = 0.00002
P(0) = [20!/0!(20-0)!] (0.5)^0 (0.5)^20 = 0.0000005
0.00002+0.0000005 = 0.0000205
Then the normal approximation for P(y<2).
P(y<2) = P(y less than or equal to 1)
convert to the standard normal z value using the correction for continuity
z = [(a+0.5) - mu]/sigma
mu = np = 20(0.5) = 10
sigma = sqre[npq] = sqre[20(0.5)(0.5)] = 2.236
z = [(1+0.5) - 10]/2.236 = -3.8
using normal curve table, my table only goes up to 3.09 so I guessed at what 3.8 would be, about 0.4998
0.5 - 0.4998 = 0.0002
P(y<2) = P(y less than or equal to 1) = P(z less than or equal to -3.8) = 0.0002
Part (b) is where I get really stumped, my textbook says to "express the binomial probability to be approximated in the form P(x less than or equal to a)." I have no idea how to do this with a decimal number.
Does P(y<2.5) = P(y less than or equal to 1.5) ??? This doesn't seem to make any sense.
Also with part (c) I don't know how to express P(3.5<y<6.5) in the correct form.
Maybe P(y less than or equal to 5.5) - P(y less than or equal to 1.5)
I'm confused, let me know what I'm doing wrong.