Question

5. The Inland Revenue department claims it takes an average of 6 hours to complete a tax form. Assuming the time to complete the form is normally distributed with a standard deviation of 45 minutes:
i. What percent of people would you expect to complete the form in more than 5 hours?


ii. If a sample of 20 taxpayers are selected, what is the probability that the sample mean is less than 6.5 hours?​

Answer 1

Answer:

i) 90.82% of the people can complete the form in more than 5 hours

ii) The probability that the sample mean is less than 6.5 hours is 0.9986 (99.86%)

Step-by-step explanation:

* Lets explain how to sole the problem

- The average time to complete a tax form is 6 hours

- Assuming the time to complete the form is normally distributed with a

 standard deviation of 45 minutes

∵ 1 hour = 60 minutes

∴ 45 minutes = 45/60 = 0.75 hour

∴ The standard deviation is 0.75 hour

i)

- We need to know the percentage of the people who complete the

 form in more than 5 hours

* Lets use z-score to find the probability

∵ z-score = (x - μ)/σ , where

# x is the score

# μ is the mean

# σ is the standard deviation

∵ μ = 6 , σ = 0.75 , x = 5

∴ $z=\frac{5-6}{0.75}=-1.33$

- Use the normal distribution table to find the corresponding area

 of z-score -1.33

∵ The corresponding area of -1.33 = 0.09176

∵ P(x > 5) = 1 - P(z > -1.33)

∴ P(x > 5) = 1 - 0.09176 = 0.90824

- Change it to percentage by multiply it by 100%

∴ P(x > 5) = 0.90824 × 100% = 90.82%

∴ 90.82% of the people can complete the form in more than 5 hours

ii)

- A sample of 20 taxpayers are selected

- We need to know the probability that the sample mean is less

 than 6.5 hours

- The sample means is called M

- The standard deviation of the distribution of sample means is

  called σM  where σM = σ/√n , where σ is the standard deviation and

  n is the sample size

- z-score = (M - μ)/σM, where μ is the mean of the population  

∵ μ = 6 , σ = 0.75 , n = 20 ,  M = 6.5 hours

∴ σM = $\frac{0.75}{\sqrt{20}}=0.1677$

∴ $z=\frac{6.5-6}{0.1677}=2.98$

- Use the normal distribution table to find the corresponding area

 of z-score 2.98

∵ The corresponding area of 2.98 is 0.99856

∵ P(x < 6.5) = P(z < 2.98)

∴ P(x < 6.5) = 0.99856

∴ The probability that the sample mean is less than 6.5 hours

   is 0.9986 (99.86%)