Question

Suppose 30% of the U.S. population has green eyes. If a random sample of size 1200 U.S. citizens is drawn, then the probability that less than 348 U.S. citizens have green eyes is _______.

Answer 1

Answer:

P(X is less than 348) = 0.2148

Step-by-step explanation:

Given that:

Sample proportion (p) = 0.3

Sample size = 1200

Let X be the random variable that obeys a binomial distribution. Then;

$X \sim Bin(n = 1200,p =0.3)$

The Binomial can be approximated to normal with:

$\mu = np = 1200 \times 0.3 \\ \\ \mu= 360$

$\sigma = \sqrt{np(1-p) } \\ \\ \sigma = \sqrt{1200 \times (0.3)(1-0.3) } \\ \\ \sigma = 15.875$

To find:

P(X< 348)

So far we are approximating a discrete Binomial distribution using the continuous normal distribution. 348 lies between 347.5 and 348.5

Normal distribution:

x = 347.5, $\mu$ = 360, $\sigma$ = 15.875

Using the z test statistics;

$z = \dfrac{x - \mu}{\sigma}$

$z = \dfrac{347.5 - 360}{15.875}$

$z = \dfrac{-12.50}{15.875}$

z = -0.7874

z ≅ - 0.79

The p-value for P(X<347.5) = P(Z < -0.79)

From the z tables;

P(X<347.5) = 0.2148

Thus;

P(X is less than 348) = 0.2148