Binomial Distribution? ?

Question

a) Given X is Binomial with with n = 11 and p = 0.4 use the Normal Approximation to the Binomial Distribution to find P( X < 5).

b) Suppose X in not nomal with MEAN = 25 and ST. DEV. 6. Based on a sample of size

n = 16 find the probability that a SAMPLE MEAN will be greater than 29.

Answer 1

a) mean =np = 11(0.4) = 4.4

variance = np(1-p)

=11(0.4)(0.6)=2.64

Standard deviation = sqrt[variance] =1.6248

Convert X to Z, where Z= (5-4.4)/1.6248=0.3693

P(X &lt; 5) = P( Z &lt; 0.3693) = 0.6443

Note: For large samples, a continuity correction is applied. Because this is a small sample, a normal approximation is not needed. So, I have skipped the continuity correction.

b)P(xbar &gt;29) =

Mean μ = 25

Standard deviation σ = 6

Standard error σ / √ n = 6 / √ 16 = 1.5

P(x &gt; 29) = P( z &gt; (29-25) / 1.5)

= P(z &gt; 2.6667) = 0.0038

Because of Central Limit Theorem, the distribution of sample means approach a Normal distribution with mean  ÃŽÂ¼ and standard deviation  ÃÂƒ / √ n, even if the original distribution is not normal..