cling tras educing aste, and reusing materials are eco-actions that ill help the
ID: 3363365 • Letter: C
Question
cling tras educing aste, and reusing materials are eco-actions that ill help the environment. Ac rding to a A Today snapshot, 78% of respondents list recycling as the leading way to help our environment. Suppose that a dom samp n = 100 adults is selected and that the 78% figure is correct. Survey say recycling trash, reducing waste, and reusing materials Rayde make the biggestune Ornic difference in | Dome. protecting the a) Does the distribution of p, the sample proportion of adults who list recycling as the y to help the environment have an approximate normal distribution? If so, what is its mean and standard deviation? Ycs, the distribution is approximately normal. The mean is 78 and the standard deviation is 22 o Yes, the distribution is approximately normal. The mean is 0.78 and the standard deviation is 0.0414. Yes, the distribution is approximately normal, The mean is 0.78 and the standard deviation is 0.22 No, the distribution is not normal because np is less than 5 No, the distribution is not normal because nq is less than 5. (b) what is the probability that the sample proportion p is less than 73%? (Round your answer to four decimal places.) 0.1131 (c)What is the probability that p lies in the interval 0.67 to 0.73 not including the endpoints? (Round your answer to four decimal places.) 0.1092Explanation / 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
a.
P(X < 0.73) = (0.73-0.78)/0.0414
= -0.05/0.0414= -1.2077
= P ( Z <-1.2077) From Standard Normal Table
= 0.1136
b.
BETWEEN THEM
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 0.67) = (0.67-0.78)/0.0414
= -0.11/0.0414 = -2.657
= P ( Z <-2.657) From Standard Normal Table
= 0.00394
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(0.67 < X < 0.73) = 0.11358-0.00394 = 0.1096
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.