Question

8.9 According to the National Center for Health Statistics, as of 2012, it is estimated that 24% of adults in Missouri smoke. You wish to know the probability of selecting from this popula- tion a random sample of 30 people containing 8 or fewer smokers. a. Use a statistical program to determine the exact probability of getting 8 or fewer smok- ers. (Hint: refer to section 5.4.) b. Now use the normal approximation of the binomial distribution to answer the same question. Compare your answers. Are they similar?

Please answer this question/explain entirely thanks!

Answer 1

(a)

Here, p = 0.24 (24%)

probability of getting 8 or few smokers out of 30,

$= \binom{30}{8} 0.24^8 * (1 - 0.24)^{30 - 8} + \binom{30}{7} 0.24^7 * (1 - 0.24)^{30 - 7} + ....$

$.....+\binom{30}{0} 0.24^0 * (1 - 0.24)^{30 - 0}$

= 0.719 (Using R studio)

### Code

r <- 0:8
p = 0.24
n = 30
q = 1 - p

sum <- 0

for (i in r) {
value <- choose(n, i) * (p ^ i) * (q ^ (n - i))
sum <- sum + value
}

(b)

If we approximate the distribution as normal distribution then if X is the random variable of that distribution such that,

$X \sim N(np, np(1-p))$

$X \sim N(30 * 0.24, 30 * 0.24*(1-0.24))$

$X \sim N(7.2, 5.472)$

Now,

$P (X \leq 8) = P(Z \leq \frac{8 - 7.2}{\sqrt {5.472}}) = P(Z \leq 0.342) = 0.6338$

Comparing we get that the result is not similar. The reason is the small sample size. As the sample size gets larger the binomial distribution tends to approximate normal distribution.

** If the answer does not match please comment.