\"What do you think is the ideal number of children for a family to have?\" A Ga
ID: 2923888 • Letter: #
Question
"What do you think is the ideal number of children for a family to have?" A Gallup Poll asked this question of 1016 randomly chosen adults. Almost half (49%) thought two children was ideal.† We are supposing that the proportion of all adults who think that two children is ideal is p = 0.49. What is the probability that a sample proportion p falls between 0.46 and 0.52 (that is, within ±3 percentage points of the true p) if the sample is an SRS of size n = 400? (Round your answer to four decimal places.) This was my answer .8552 Incorrect: Your answer is incorrect. What is the probability that a sample proportion p falls between 0.46 and 0.52 if the sample is an SRS of size n = 5000? (Round your answer to four decimal places.) .9482 Incorrect: Your answer is incorrect.
Explanation / Answer
NORMAL DISTRIBUTION
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 ) = 0.49
standard Deviation ( sd )= sqrt(PQ/n) = sqrt(0.49*0.51/400)
=0.025
a.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 0.46) = (0.46-0.49)/0.025
= -0.03/0.025 = -1.2
= P ( Z <-1.2) From Standard Normal Table
= 0.11507
P(X < 0.52) = (0.52-0.49)/0.025
= 0.03/0.025 = 1.2
= P ( Z <1.2) From Standard Normal Table
= 0.88493
P(0.46 < X < 0.52) = 0.88493-0.11507 = 0.7699
b.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 0.46) = (0.46-0.49)/0.0070696535
= -0.03/0.0070696535 = -4.243489444
= P ( Z <-4.243489444) From Standard Normal Table
= 0.00001
P(X < 0.52) = (0.52-0.49)/0.0070696535
= 0.03/0.0070696535 = 4.243489444
= P ( Z <4.243489444) From Standard Normal Table
= 0.99999
P(0.46 < X < 0.52) = 0.99999-0.00001 = 0.9999779929 ~ 1
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.