Question

Q1. [5 pts] Suppose that Z follows a standard normal. Calculate the probability that:

a)Z is less than 1.25.

b)Z is greater than 1.75.

c)Z is between 1.25 and 1.75.

d)Z is between -1.25 and 1.75.

e)X is between 85 and 105, where X follows a normal with mean of 100 and SD of 10.Note: Write probabilities as decimals and to the four decimal places (e.g., 0.2538).

Q2. What is the value of z such that:a)the probability of a standard normal variable exceeding z is 55%.

b)the probability of a standard normal variable exceeding z is 75%.

c)the probability of a standard normal variable being less than z is 30%

d)the probability of a standard normal variable being less than z is 60%

e)the interval [-z, z] contains 70% of all possible z values.Note: Write numbers to the three decimal places (e.g., 1.964).

Q3. The management of a local restaurant wants to determine the average monthly expenditure by households in fancy restaurants. Management wants to be 90% confident of the findings and does not want the error to exceed ±$5.

a)Past studies indicate that the population SD is $30. What sample size should be used?

b)Suppose that the management collected a sample of the size calculated in

a). It finds in the sample that the average expenditure was $80 and SD was $35. Construct a 90% CI for the true expenditure.

Note: Write the lower and upper bounds in the CI to the two decimal places (e.g., [11.33, 15.55]).c)If someone insists that the true monthly expenditure is $73. Based your answer to b), would you agree or disagree with this person? Why agree or why not agree?

Q4. [10 pts] To determine the effectiveness of the advertising campaign for a new digital video recorder, management would like to know what proportionof the households is aware of the brand. The advertising agency thinks that this figure is close to .55. The management would like to have a margin of error of ±.025 at the 99% confidence level.

a)What sample size should be used?

b)A sample of the size calculated in a) has been taken. The management found the sample proportion to be .575. Construct a 99% CI for the true proportion.

Note: Write the lower and upper bounds in the CI to the four decimal places (.e.g, [.3012, .3875]).c)If someone insists that the true proportion is .59. Based your answer to b), would you agree or disagree with this person? Why agree or why not agree?

Answer 1

Solution:-

Mean = 100, S.D = 10

1)

a) The probability that z is less than 1.25 is 0.894.

P(z < 1.25) = 0.894

b) The probability that z is greater than 1.75 is 0.04.

P(z > 1.75) = 0.04

c) The probability that z is between 1.25 and 1.75 is 0.066.

P( 1.25 < z < 1.75) = P(z > 1.25) - P(z > 1.75)

P( 1.25 < z < 1.75) = 0.106 - 0.04

P( 1.25 < z < 1.75) = 0.066

d) The probability that z is between -1.25 and 1.75 is 0.854.

P(-1.25 < z < 1.75) = P(z > -1.25) - P(z > 1.75)

P(-1.25 < z < 1.75) = 0.894 - 0.04

P(-1.25 < z < 1.75) = 0.854

e) The probability that X is between 85 and 105 is 0.624.

x₁ = 85

x₂ = 105

By applying normal distribution:-

z₁ = - 1.50

z₂ = 0.50

P(-1.50 < z < 0.50) = P(z > -1.50) - P(z > 0.50)  

P(-1.50 < z < 0.50) = 0.933 - 0.309

P(-1.50 < z < 0.50) = 0.624