Answer to Question #139707 in Statistics and Probability for Lin

Question

QUESTION B1

A study of corruption for a certain geographic region showed an average of 5 corruptions occur per 20,000 people. In a city of 80,000 people, find the probability that at least 3 corruptions occur.

QUESTION B2

(a) A random sample of 300 students from a geographic area indicated that 45 students attended private school. Estimate the true proportion of students attending private schools with 95% confidence.

(b) A survey of 25 individuals who passed the ACCA exam showed that average annual
salary is RM150,000 and standard deviation RM15,000.

(i) Find the degrees of freedom.

(ii) Construct 99% confidence interval for the true average salary.

Answer 1

QUESTION B1

Let $X=$the number of corruption occuried : $X\sim Po(\lambda)$

$$P(X=x)=\dfrac{e^{-\lambda}\lambda^x}{x!}$$

Given

$$\lambda=5\cdot\dfrac{80000}{20000}=20$$

$$P(X\geq3)=1-(P(X=0)+P(X=1)+P(X=2))=$$

$$=1-\dfrac{e^{-20}\cdot 20^0}{0!}-\dfrac{e^{-20}\cdot 20^1}{1!}-\dfrac{e^{-20}\cdot 20^2}{2!}=$$

$$=1-e^{-20}(1+20+200)=1-221\cdot e^{-20}\approx$$

$$\approx0.99999954$$

QUESTION B2

(a)

Favorable Cases $x=45$  

Sample Size $n=300$  

The sample proportion $\hat{p}=\dfrac{x}{n}=\dfrac{45}{300}=0.15$

The critical value for $\alpha=0.05$ is $z_c=1.96$. The corresponding confidence interval is computed as shown below:

$$CI(Proportion)=(\hat{p}-z_c\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}, \hat{p}+z_c\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}})=$$ $$=(0.15-1.96\sqrt{\dfrac{1.5(1-1.5)}{300}}, 0.15+1.96\sqrt{\dfrac{0.15(1-0.15)}{300}})=$$

$$=(0.1096, 0.1904)$$

Therefore, based on the data provided, the 95% confidence interval for the population proportion is $0.1096<p<0.1904,$ which indicates that we are 95% confident that the true population proportion $p$ is contained by the interval $(0.1094, 0.1904).$

b)

(i)

$df=n-1=25-1=24$

(ii) Based on the provided information, the critical t-value for $\alpha=0.01$ and $df=24$

degrees of freedom is $t_c=2.79694.$

The 99% confidence for the population mean $\mu$ is computed using the following expression

$$CI=(\bar{x}-t_c\times\dfrac{s}{\sqrt{n}}, \bar{x}+t_c\times\dfrac{s}{\sqrt{n}})=$$

$$=(150000-2.79694\times\dfrac{15000}{\sqrt{25}}, 150000+2.79694\times\dfrac{15000}{\sqrt{25}})=$$

$$=(141609,158391)$$