A fair coin is tossed 10 times. Find the probability of getting at least 4 heads and at most 6 heads.
Let X be the probability distribution of getting x heads. We need to find $k$ such that $k = P(4 \le X \le 6)$.
It is a binomial distribution with n = 10, p = q = 0.5. Then, k = B(6, 10, 0.5) – B( 3 , 10, 0.5) = 0.656 (from table). This is because, for this discrete distribution, we have to subtract the values of P(X = 0, 1, 2, 3) from B(6, 10, 0.5).
When I approximate it by normal distribution N(5, 2.5), I found that I have to use $X_1 = 3$ (instead of 4) and $X_2 = 6$ to get the respective standard scores $Z_1 = –1.265$ and $Z_2 = 0.633$. The corresponding area under the standard normal curve is then 0.634 giving a very close approximate to the result previously found.
My question is:- using $X_1 = 3$ and $X_2 = 6$ to find the corresponding area under the normal curve will include those X = 3.xxxxx ~ X = 4.0. Shouldn't those areas be excluded according to the question?