3)20% of ACME's employees are senior citizens. What is the probability that in a random sample of 100 employees, that the sample proportion of senior citizens will be less than 15%? Provide a diagram.

4)ROYAL FINANCE takes a sample of 5 employees and finds that the sample mean hourly rate was $175 with a sample standard deviation of $30. Construct a 90% confidence interval for the population average hourly rate of all Royal Finance employees. 

Q3)The probability that in a random sample of 100 employees, that the sample proportion of senior citizens will be less than 15% is 0.10565

90% confidence interval is (146.40,  203.60)

Step-by-step explanation

Q3. 

p = 0.20

q = 1 - 0.20 = 0.80

n = 100

p_hat = 0.15

Z = (p_hat - p)/√(p*q/n) 

Z = (0.15 - 0.20)/√(0.20*0.80/100)

Z = - 1.25

P(Z < - 1.25) = 0.10565

Q4. 

Use t distribution since the population standard deviation is unknown. 

n = 5 

Mean = 175

Standard deviation = 30

For 90% confidence interval, 

Alpha/2 = (1 - 0.90)/2 

               = 0.05

Degrees of freedom, df = n - 1 

                                      df = 5 - 1 =4 

Using df = 4 and alpha/2 = 0.05, 

t_critical = 2.132 (from the t distribution table) 

Margin of error = t_critical *(s/√n) 

                            = 2.132*(30/√5)

                            = 28.6038

Lower limit = mean - Margin of error 

                   = 175 - 28.6038

                   = 146.40

Upper limit = mean + Margin of error 

                    = 175 + 28.6038

                    = 203.60

90% confidence interval is (146.40  203.60)

Attached is the diagram 

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