A college research group reported that 47% of college students aged 18-24 would

Question

A college research group reported that 47% of college students aged 18-24 would spend their spring breaks relaxing at home in 2009 A sample of 160 college students was selected. Complete parts a through d below. a. Calculate the standard error of the proportion p = L (Round to four decimal places as needed.) b. What is the probability that less than 40% of the college students from the sample spent their spring breaks relaxing at home? P(Less than 40% of the college students from the sample spent their spring breaks relaxing at home) (Round to four decimal places as needed.)

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 ) = 0.47
a.
standard Deviation ( sd )= sqrt(PQ/n) = sqrt(0.47*0.53/160)
=0.0395
b.
LESS THAN
P(X < 0.4) = (0.4-0.47)/0.0395
= -0.07/0.0395= -1.7722
= P ( Z <-1.7722) From Standard Normal Table
= 0.0382
c.
P(X > 0.5) = (0.5-0.47)/0.0395
= 0.03/0.0395 = 0.7595
= P ( Z >0.759) From Standard Normal Table
= 0.2238
d.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 0.48) = (0.48-0.47)/0.0395
= 0.01/0.0395 = 0.2532
= P ( Z <0.2532) From Standard Normal Table
= 0.59993
P(X < 0.58) = (0.58-0.47)/0.0395
= 0.11/0.0395 = 2.7848
= P ( Z <2.7848) From Standard Normal Table
= 0.99732
P(0.48 < X < 0.58) = 0.99732-0.59993 = 0.3974

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