5:25 a wamap.org According to a 2009 Reader\'s Digest article, people throw away

Question

5:25 a wamap.org According to a 2009 Reader's Digest article, people throw away approximately 13% of what they buy at the grocery store. Assume this is the true proportion and you plan to randomly survey 205 grocery shoppers to investigate their behavior. What is the probability that the sample proportion is betweern 0.07 and 0.13? Note: You should carefully round any intermediate values you calculate to 4 decimal places to match wamap's approach and calculations. Answer (Enter your answer as a number accurate to 4 decimal places.) License Points possible: 10 This is attempt 1 of 5. Post this question to forum Submit

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 ) = 0.13
standard Deviation ( sd )= sqrt(PQ/n) = sqrt(0.13*0.87/205)
=0.0235

To find P(a < = Z < = b) = F(b) - F(a)
P(X < 0.07) = (0.07-0.13)/0.0235
= -0.06/0.0235 = -2.5532
= P ( Z <-2.5532) From Standard Normal Table
= 0.00534
P(X < 0.13) = (0.13-0.13)/0.0235
= 0/0.0235 = 0
= P ( Z <0) From Standard Normal Table
= 0.5
P(0.07 < X < 0.13) = 0.5-0.00534 = 0.4947

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