Recycling trash, reducing waste, and reusing materials are eco-actions that will

Question

Recycling trash, reducing waste, and reusing materials are eco-actions that will help the environment. According to a USA Today snapshot, 78% of respondents list recycling as the leading way to help our environment. Suppose that a random sample of n = 100 adults is selected and that the 78% figure is correct.

(b) What is the probability that the sample proportion p is less than 73%? (Round your answer to four decimal places.)

(c) What is the probability that p lies in the interval 0.66 to 0.73 not including the endpoints? (Round your answer to four decimal places.)

(d) What might you conclude about p if the sample proportion were less than 0.65? (Choose 1 from below)

The value p = 0.65 is between 2 and 3 standard deviations from the mean. Therefore, it is an unlikely occurrence if p = 0.78. Perhaps the sampling was not random, or the 78% figure is not correct.

The value p = 0.65 is more than 3 standard deviations from the mean. Therefore, it is a likely occurrence if p = 0.78. There is no evidence of an inaccurate value of p.

The value p = 0.65 is more than 3 standard deviations from the mean. Therefore, it is an unlikely occurrence if p = 0.78. Perhaps the sampling was not random, or the 78% figure is not correct.

The value p = 0.65 is less than 1 standard deviation from the mean. Therefore, it is a likely occurrence if p = 0.78.

There is no evidence of an inaccurate value of p.The value p = 0.65 is between 2 and 3 standard deviations from the mean. Therefore, it is an unlikely occurrence if p = 0.78. There is no evidence of an inaccurate value of p.

Explanation / Answer

the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
proportion ( p ) = 0.78
standard Deviation ( sd )= sqrt(PQ/n) = sqrt(0.78*0.22/100)
=0.0414
b.
P(X > 0.73) = (0.73-0.78)/0.0414
= -0.05/0.0414 = -1.2077
= P ( Z >-1.208) From Standard Normal Table
= 0.8864
c.
BETWEEN THEM
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 0.66) = (0.66-0.78)/0.0414
= -0.12/0.0414 = -2.8986
= P ( Z <-2.8986) From Standard Normal Table
= 0.00187
P(X < 0.73) = (0.73-0.78)/0.0414
= -0.05/0.0414 = -1.2077
= P ( Z <-1.2077) From Standard Normal Table
= 0.11358
P(0.66 < X < 0.73) = 0.11358-0.00187 = 0.1117
d.
P(X < 0.65) = (0.65-0.78)/0.0414
= -0.13/0.0414= -3.1401
= P ( Z <-3.1401) From Standard Normal Table
= 0.0008
The value p = 0.65 is more than 3 standard deviations from the mean. Therefore, it is a likely occurrence if p = 0.78. There is no evidence of an inaccurate value of p.

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