Question

Use the follow information to answer questions 1-5.

In a restaurant, the proportion of people who order coffee with their dinner is 0.75. A simple random sample of 200 patrons of the restaurant is taken.

Calculate the expected value of the sample proportion.

	a.	0.25
	b.	0.37
	c.	0.75

Calculate the standard deviation of the sample proportion.

	a.	0.01
	b.	0.03
	c.	0.25

Using the two conditions, np ≥ 5, and n (1 - p) ≥ 5, can the sampling distribution of p be approximated by a normal distribution?

	a.	Yes
	b.	No

Compute the probability that the proportion of people who will order coffee with their meal is at most 0.8.

	a.	0.80
	b.	0.05
	c.	0.95

Compute the probability that the proportion of people who will order coffee with their meal is at least 0.8.

	a.	0.05
	b.	0.95
	c.	0.80

Answer 1

Solution :

Given that p = 0.75 , n = 200

=> q = 1 - p = 0.25

=> Answer :- Option c. 0.75

=> The expected value of the sample proportion is E(x) = μp^ = p = 0.75

=> Answer :- Option b. 0.03

=> The standard deviation of the sample proportion is = σp^ =

= sqrt(p*q/n)

= sqrt(0.75*0.25/200)

= 0.0306

= 0.03 (rounded)

=> Answer :- Option a. Yes
  
=> n*p = 200*0.75 = 150 >= 5 and
  
n*(1 - p) = 200*0.25 = 50 >= 5

=> Therefore we can conclude that sampling distribution of p is approximately a normal distribution with mean μp^ = 0.75 and standard deviation σp^ = 0.03

=> Answer :- Option c. 0.95

=> P(p^ <= 0.8) = P((p^ - μp^)/σp^ <= (0.8 - 0.75)/0.03)

= P(Z <= 1.6667)

= 0.9525

= 0.95 (rounded)

=> Answer :- Option a. 0.05

=> P(p^ >= 0.8) = P((p^ - μp^)/σp^ >= (0.8 - 0.75)/0.03)

= P(Z >= 1.6667)

= 1 - P(Z < 1.6667)

= 1 - 0.9525

= 0.0475

= 0.05 (rounded)