# Solved (Free): According to the 2015 national health interview survey, 46.9% persons with asthma having asthma attacks

#### ByDr. Raju Chaudhari

Mar 10, 2021

According to the 2015 national health interview survey, 46.9% persons with asthma having asthma attacks during the previous 12 months.

(1) What is the probability that, among ten patients with asthma, five or more will have asthma attacks during the previous 12 months?

(2) What is the probability that, among 20 patients with asthma, at most six will have asthma attacks during the previous 12 months?

#### Solution

Let $p=0.469$ be the probability that patients with asthma having attack.

Let $n = 10$, number of patients with asthma. Let $X$ denote the number of patients with asthma having attack.

$X\sim B(n,p)$. That is $X\sim B(10, 0.469)$.

The probability mass function of $X$ is

 \begin{aligned} P(X=x) &= \binom{10}{x} (0.469)^x (1-0.469)^{10-x},\\ &\quad x=0,1,\cdots, 10 \end{aligned}

(1) The probability that, among ten patients with asthma, five or more will have asthma attacks during the previous 12 months is

 \begin{aligned} P(X\geq 5) &= 1- P(X < 5)\\ & =1- P(X\leq 4)\\ &= 1- \sum_{x=0}^{4} P(X=x)\\ &= 1-\bigg(P(X=0) + P(X=1)+P(X=2)\\ &\quad +P(X=3)+P(X=4)\bigg)\\ & = 1- \bigg[\binom{10}{0} (0.469)^0 (1-0.469)^{10-0} + \binom{10}{1} (0.469)^1 (1-0.469)^{10-1}\\ &\quad +\binom{10}{2} (0.469)^2 (1-0.469)^{10-2} \\ & \quad \binom{10}{3} (0.469)^3 (1-0.469)^{10-3}+\binom{10}{4} (0.469)^4 (1-0.469)^{10-4}\bigg]\\ &=1- (0.0018+0.0157+0.0626+0.1474+0.2278)\\ &= 1- 0.4552\\ & = 0.5448 \end{aligned}

(2) The probability that, among 20 patients with asthma, at most six will have asthma attacks during the previous 12 months

$X\sim B(n,p)$. That is $X\sim B(20, 0.469)$.

The probability mass function of $X$ is

 \begin{aligned} P(X=x) &= \binom{20}{x} (0.469)^x (1-0.469)^{20-x},\\ &\quad x=0,1,\cdots, 20 \end{aligned}

The probability that, among 20 patients with asthma, at most six will have asthma attacks during the previous 12 months

 \begin{aligned} P(X\leq 6) &= \sum_{x=0}^{6} P(X=x)\\ &= \bigg(P(X=0) + P(X=1)+P(X=2)+P(X=3)\\ &\quad +P(X=4)+P(X=5)+P(X=6)\bigg)\\ & = \bigg[\binom{20}{0} (0.469)^0 (1-0.469)^{20-0} + \binom{20}{1} (0.469)^1 (1-0.469)^{20-1}+\binom{20}{2} (0.469)^2 (1-0.469)^{20-2} \\ & \quad \binom{20}{3} (0.469)^3 (1-0.469)^{20-3}+\binom{20}{4} (0.469)^4 (1-0.469)^{20-4}\\ & \quad \binom{20}{5} (0.469)^5 (1-0.469)^{20-5}+\binom{20}{6} (0.469)^6 (1-0.469)^{20-6}\bigg]\\ &=(0+0.0001+0.0005+0.0025+0.0094+0.0265+0.0584)\\ &= 0.0973 \end{aligned}