Answer to Question #232913 in Statistics and Probability for Kudzie

Question

The average dividend yield of a random sample of 25 JSE-listed companies this year was found 

to be 14.5%, with a sample standard deviation of 3.4%. Assume that dividend yields are 

normally distributed. 

3.1.1 Calculate, with 90% confidence, the actual mean dividend yield of all JSE-listed 

companies this year. Interpret the finding. (6)

3.1.2 Calculate, with 95% confidence, the actual mean dividend yield of all JSE-listed 

companies this year. Compare the interval with the one calculated in 3.1.1. 

 (12)

3.2 Norman is a student at a college in Durban. The amount of time, in minutes, that Norman 

walks to the college for his final examinations is constantly distributed between 15 to 40 

minutes, inclusive. Use this information to answer the following questions.

3.2.1 Name the continuous probability distribution described above. Explain in detail why it 

is called the distribution of little information. (3

Answer 1

3.1.1

We are given that the sample mean = 0.145 and the sample standard deviation, s = 0.034 for sample size n=25

The Z value corresponding to 90% CI = 1.645

The 90% CI = mean ± Z*s/$\sqrt{\smash[b]{n}}$

The lower bound = 0.145 - 1.645*0.034/$\sqrt{\smash[b]{25}}$  = 0.1338

The upper bound = 0.145 + 1.645*0.034/$\sqrt{\smash[b]{25}}$  = 0.1562

The 90% CI = (0.1338, 0.1562)

This means that we are 90% confident that the actual mean dividend yield of all JSE-listed companies this year will be contained in the interval from 13.38% to 15.62%

3.1.2

We are given that the sample mean = 0.145 and the sample standard deviation, s = 0.034 for sample size n=25

The Z value corresponding to 95% CI = 1.96

The 95% CI = mean ± Z*s/$\sqrt{\smash[b]{n}}$

The lower bound = 0.145 - 1.96* 0.034/$\sqrt{\smash[b]{25}}$ = 0.1317

The upper bound = 0.145 + 1.96*0.034/$\sqrt{\smash[b]{25}}$  = 0.1583

The 95% CI = (0.1317, 0.1583)

This means that we are 95% confident that the actual mean dividend yield of all JSE-listed companies this year will be contained in the interval from 13.17% to 15.83%

We find that the 95% confidence level has a wider interval than the 90% confidence interval. This may imply that the more the confidence level, the wider the interval.

3.2.1

This is the uniform distribution since the probability is constant between 15 and 40 minutes.

3.2.2

= P(20<x<38)

probability = (38-28)/(40-15)

= 10/25

= 0.4