Question

[4] Approximate the following binomial probabilities by the use of normal approximation (n = 50, p = 0.3). Remember to use a continuity correction. P(x = 18) P(x ≥ 15) P(x ≤ 12) P(12 ≤ x ≤ 18)

I don't understand this question as well too?

Answer 1

Normal approximation is used in binomial probability when the 'n' is too large ( < 30) and when np > > 5. We do it since it becomes tedious to calculate when 'n' increases. Since it is approximation the answer will be close to the correct answer. Also we know that binomial is discrete and normal is continuous hence we need to do the continuity correction. We do this because we calculate the discrete at fixed intervals whereas the probability of continuous at a point is '0'.

Continuous P(X = x) = 0

Hence when approximating from discrete to continuous we need to find limits or range of X.

Since E(X) = np V(X) = np(1-p)

We use normal probability tables to calculate the probabilities.

For continuity correction we can add or less 0.5.

P(X=n) = P(n – 0.5 < X < n + 0.5)

P(X > n) = P(X > n + 0.5)

P(X ≤ n) = P(X < n + 0.5)

P (X < n) = P(X < n – 0.5)

P(X ≥ n) =  P(X > n – 0.5)

P(X=18). That means X is strictly between 17 and 19

=

= P (0.77 < Z < 1.08)

= P( Z < 1.08) - P( Z < 0.77)

= 0.86 - 0.78

P(X ≥ 15) = P( X > 14.5)

= P( Z > -0.15)

= P( Z < 0.15)

P(X ≤ 12) = P(X < 12.5)

= P ( Z < -0.77)

= 1 - P( Z < 0.77)

= 1 - 0.78

P(12 ≤ X ≤ 18) = P( 12.5 <X < 17.5)

= P( X < 17.5) - P( X < 12.5)

= P( Z < 0.77) - P( Z < -0.77)

= P( Z < 0.77) - [1 - P(Z < 0.77) ]

= 2 P(Z <0.77) - 1

= 2 * 0.78 - 1