Question

A national caterer determined that 36% of the people who sampled their food said that it was delicious. A random sample of n=144 people is obtained from a large population. The 144 people are asked to sample the caterer's food. What is the probability that more than 34% thought the food was delicious?

Question 20 options:

	A) 0.6915
	B) 0.5987
	C) 0.3085
	D) 0.4013

Question 19 (1 point)

A national caterer determined that 36% of the people who sampled their food said that it was delicious. A random sample of n=144 people is obtained from a large population. The 144 people are asked to sample the caterer's food. What is the probability that less than 32% thought the food was delicious?

Question 19 options:

	A) 0.8413
	B) 0.1587
	C) 0.4602
	D) 0.3174

Question 18 (1 point)

A national caterer determined that 36% of the people who sampled their food said that it was delicious. A random sample of n=144 people is obtained from a large population. The 144 people are asked to sample the caterer's food. If p̂ {"version":"1.1","math":"\(\hat{p}\)"} is the sample proportion saying that the food is delicious, what is the standard deviation of the sampling distribution of p̂ {"version":"1.1","math":"\(\hat{p}\)"}? In other words, what is σp̂ {"version":"1.1","math":"\(\sigma_\hat{p}\)"}?

Question 18 options:

	A) 0.48
	B) 0.04
	C) 0.003333
	D) 5.76

Answer 1

Solution

Given that,

p = 0.36

1 - p = 1 - 0.36 = 0.64

n = 144

= p = 0.36

=  [p( 1 - p ) / n] = [(0.36 * 0.64) / 144 ] = 0.04

1) P( > 0.34) = 1 - P( < 0.34 )

= 1 - P(( - ) / < (0.34 - 0.36) / 0.04)

= 1 - P(z < -0.50)

Using z table

= 1 - 0.3085

= 0.6915

correct option is = A

2) P( < 0.32)

= P[( - ) / < (0.32 - 0.36) / 0.04]

= P(z < -1.00)

Using z table,

= 0.1587

correct option is = B

3) =  [p( 1 - p ) / n] = [(0.36 * 0.64) / 144 ] = 0.04

correct option is = B