Answer to Question #294789 in Statistics and Probability for John

Question

A random sample of size 50 is drawn from binomial distribution with parameters n=100 and p=0.3. what is the probability that the sample mean is

a)larger than 3.9

b) between 4.1 and 4.4

c) smaller than 4.0

Answer 1

Let $X$ be a binomial rv based on $n$ trials with success probability $p.$ Then if the

binomial probability histogram is not too skewed, $X$ has approximately a

normal distribution with $\mu=np$ and $\sigma=\sqrt{npq}.$

In practice, the approximation is adequate provided that both $np\ge 10$ and $nq\ge 10.$

Given $n=100, p=0.3, q=1-p=1-0.3=0.7.$

$np=100(0.3)=30\ge 10$

$nq=100(0.7)=70\ge 10$

We can use normal approximation for binomial distribution with $\mu=np=100(0.3)=30, \sigma=\sqrt{npq}=\sqrt{100(0.3)(0.7)}=\sqrt{21}$

Let $\bar{X}=$ the sample mean: $\bar{X}\sim N(\mu, \sigma^2/n_1)$

Given $n_1=50.$

a)

$$P(\bar{X}>3.9)=1-P(Z\le\dfrac{3.9-30}{\sqrt{21}/\sqrt{50}})$$

$$\approx1-P(Z\le-40.2732)\approx1$$

b)

$$P(4.1<\bar{X}<4.4)=P(Z<\dfrac{4.4-30}{\sqrt{21}/\sqrt{50}})$$

$$-P(Z\le\dfrac{4.1-30}{\sqrt{21}/\sqrt{50}})$$

$$\approx P(Z<-39.5017)-P(Z\le-39.9646)\approx0$$

c)

$$P(\bar{X}<4.0)=P(Z<\dfrac{4.0-30}{\sqrt{21}/\sqrt{50}})$$

$$\approx P(Z<-40.1189)\approx0$$