Q. 5.20 - A First Course in Probability

Question

If $65$percent of the population of a large community is in favor of a proposed rise in school taxes, approximate the probability that a random sample of $100$people will contain

(a) at least $50$who are in favor of the proposition;

(b) between $60$and $70$inclusive who are in favor;

(c) fewer than $75$ in favor.

Step-by-Step Solution

Answer

(a) The probability that at least $50$ are in favor is $0.9994$.

(b) The probability that between $60$ and $70$inclusive are in favor is $0.7499$.

(c) The probability that fewer than $75$ are in favor is $0.9767$

Part (a) Step 1. Given Information.

Here, it is given that $65\%$of population is in favor of proposed hike in school taxes.

Part (a) Step 2. Find μ   and   σ .

Let the sample size be $n=100$ people.

Let the proportion of people in the favor of the proposed hike in the school taxes be $p=0.65$.

Requirements for the normal approximation to the binomial distribution.

$$\begin{gathered} np=100\times 0.65 \\ =65 >10 \\ \text{And} \\ n\left(1-p\right)= 100\left(1-0.65\right) \\ =0.35>10 \end{gathered}$$

Hence, both requirements are satisfied because $np$and $np\left(1-p\right)$are greater than 10.

The mean and standard deviation of the binomial distribution is,

$$\begin{gathered} \mu =np \\ =100\times 0.65 \\ =65 \\ And, \\ \sigma =\sqrt{np\left(1-p\right)} \\ =\sqrt{100\times 0.65\times \left(1-0.65\right)} \\ =4.769696 \end{gathered}$$

Part (a) Step 3. Find that at least 50 are in favor of the proposition.

Let $X$be the no of people who favor proposed rise in school taxes.

Using continuity correction, the required probability is,

$$\begin{gathered} p\left(X\ge 50\right)= p \left(\frac{\left(X-0.5\right)-np}{\sqrt{np\left(1-p\right)}}\ge \frac{\left(50-0.5\right)-65}{4.769696}\right) \\ =p\left(Z\ge \frac{\left(50-0.5\right)-65}{4.769696}\right) \\ =p\left(Z\ge -3.25\right) \\ =1- p\left(Z\ge -3.25\right) \\ =1-0.0006 \\ =0.9994 \end{gathered}$$

Therefore, the probability that at least $50$ are in favor of proposition is $0.9994$.

Part (b) Step 1. Compute probability that X is between 60   a n d   70 inclusive who are in favor.

Using continuity correction, the required probability is,

$$\begin{gathered} p\left(60\le X\le 70\right) \\ =p\left[\frac{\left(60-0.5\right)-65}{4.77}\le \frac{X-np}{\sqrt{np\left(1-p\right)}}\le \frac{\left(70+0.5\right)-65}{4.77}\right] \\ =p\left(-1.15<Z<-1.15\right) \\ =p\left(Z\le 1.15\right)-p \left(Z\le -1.15\right) \\ =0.8749-0.1251 \\ =0.7499 \end{gathered}$$

Therefore, the probability that $X$ is between $60 and 70$ inclusive who are in favor is $0.7499$.

Step 5. Compute probability that X is fewer than 75 in favor

Using continuity correction the required probability is,

$$\begin{gathered} p\left(X\le 75\right) \\ =p\left[\frac{X-np}{\sqrt{np\left(1-p\right)}}<\frac{75-0.5-65}{4.77}\right] \\ =p\left(Z\le 1.99\right) \\ =0.9767 \end{gathered}$$

Therefore, the probability that $X$ is fewer than $75$ in favor is $0.9767$