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=the number of corruption occuried : X∼Po(λ)
P(X=x)=x!e−λλx
Given
λ=5⋅2000080000=20
P(X≥3)=1−(P(X=0)+P(X=1)+P(X=2))=
=1−0!e−20⋅200−1!e−20⋅201−2!e−20⋅202=
=1−e−20(1+20+200)=1−221⋅e−20≈
≈0.99999954
QUESTION B2
(a)
Favorable Cases x=45
Sample Size n=300
The sample proportion p^=nx=30045=0.15
The critical value for α=0.05 is zc=1.96. The corresponding confidence interval is computed as shown below:
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 α=0.01 and df=24
degrees of freedom is tc=2.79694.
The 99% confidence for the population mean μ is computed using the following expression
Comments
Leave a comment