find the probability of getting between 3 and 6 heads inclusive in 10 tosses of a fair coin by using
a the binomial distribution
b normal approximation to the binomial distribution
thanks for any help
btw i have 0.205 for part a
we already addressed this in another one of your posts recently. you want P(3) + P(4) + P(5) + P(6)
how is your knowledge of the binomial distribution? i can tell you $\displaystyle Bin(n,p) \approx Norm(np,npq)$ does that help? (n,p and q have the same definition as in part (a))b normal approximation to the binomial distribution
knowledge is very hazy, used to be extremely proficient in the probability area but havent been near it in a long time so finding it a bit tricky
for part a i have 0.773 happy enough with that
for part b i have the formula z = (r-Np)/(npq)^0.5
should i just apply this, i have done so and expected the two answers to correlate somewhat but they arent