Answer to Question #139707 in Statistics and Probability for Lin

Question #139707
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.
1
Expert's answer
2020-10-26T19:33:19-0400

QUESTION B1

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


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

Given

λ=58000020000=20\lambda=5\cdot\dfrac{80000}{20000}=20

P(X3)=1(P(X=0)+P(X=1)+P(X=2))=P(X\geq3)=1-(P(X=0)+P(X=1)+P(X=2))=

=1e202000!e202011!e202022!==1-\dfrac{e^{-20}\cdot 20^0}{0!}-\dfrac{e^{-20}\cdot 20^1}{1!}-\dfrac{e^{-20}\cdot 20^2}{2!}=

=1e20(1+20+200)=1221e20=1-e^{-20}(1+20+200)=1-221\cdot e^{-20}\approx

0.99999954\approx0.99999954

QUESTION B2

(a)

Favorable Cases x=45x=45  

Sample Size n=300n=300  

The sample proportion p^=xn=45300=0.15\hat{p}=\dfrac{x}{n}=\dfrac{45}{300}=0.15

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


CI(Proportion)=(p^zcp^(1p^)n,p^+zcp^(1p^)n)=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}})==(0.151.961.5(11.5)300,0.15+1.960.15(10.15)300)==(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)=(0.1096, 0.1904)

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


b)

(i)

df=n1=251=24df=n-1=25-1=24

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

degrees of freedom is tc=2.79694.t_c=2.79694.

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


CI=(xˉtc×sn,xˉ+tc×sn)=CI=(\bar{x}-t_c\times\dfrac{s}{\sqrt{n}}, \bar{x}+t_c\times\dfrac{s}{\sqrt{n}})=

=(1500002.79694×1500025,150000+2.79694×1500025)==(150000-2.79694\times\dfrac{15000}{\sqrt{25}}, 150000+2.79694\times\dfrac{15000}{\sqrt{25}})=

=(141609,158391)=(141609,158391)




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