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
Mr F says: Using my TI-89 I get 0.0000200272.
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
Mr F says: Your tables should go up to 3.8 .... But the values might not be too reliable. Using my TI-89 I get 0.00007237.
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.
Mr F says: It's not immediately clear to me why the normal approximation to Pr(y < 2) is improved by using Pr(y<2.5) rather than Pr(y < 1.5) .... Perhaps if you draw the binomial histogram and overlay the normal distribution onto it .... Maybe you could play around with this http://www.ruf.rice.edu/~lane/stat_sim/binom_demo.html.
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.
Thanks