91Ó°ÊÓ

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

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.

localid="1646656701080" $$\begin{gathered} np=100×0.65 \\ =65>10 \\ \text{And} \\ n\left(1-p\right)=100\left(1-0.65\right) \\ =0.35>10 \end{gathered}$$

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,

$$\begin{gathered} \mathrm{μ}=np \\ =100×0.65 \\ =65 \\ And, \\ \mathrm{σ}=\sqrt{np\left(1-p\right)} \\ =\sqrt{100×0.65×\left(1-0.65\right)} \\ =4.769696 \end{gathered}$$

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

Using continuity correction, the required probability is,

localid="1646738909309" $$\begin{gathered} p\left(Xâ‰\yen 50\right)=p\left(\frac{\left(X-0.5\right)-np}{\sqrt{np\left(1-p\right)}}â‰\yen \frac{\left(50-0.5\right)-65}{4.769696}\right) \\ =p\left(Zâ‰\yen \frac{\left(50-0.5\right)-65}{4.769696}\right) \\ =p\left(Zâ‰\yen -3.25\right) \\ =1-p\left(Zâ‰\yen -3.25\right) \\ =1-0.0006 \\ =0.9994 \end{gathered}$$

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

Using continuity correction, the required probability is,

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

Therefore, the probability that $X$is between $60and70$inclusive who are in favor is $0.7499$.

Using continuity correction the required probability is,

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

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