A social media survey found that 74% of parents are \"friends\" with their child

Question

A social media survey found that 74% of parents are "friends" with their children on a certain online networking site. A random sample of 150 parents was selected. Complete parts a through d below. a. Calculate the standard error of the proportion = | | (Round to four decimal places as , needed.) b. What is the probability that 118 or more parents from this sample are "friends" with their children on this online networking site? P(118 or more parents from this sample are "friends" with their children) (Round to four decimal places as needed.)

Explanation / Answer

NORMAL APPROXIMATION TO BINOMIAL 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
a.
mean ( np ) = 150 * 0.74 = 111
standard deviation ( npq )= 150*0.74*0.26 = 5.3722
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
b.
P(X < 118) = (118-111)/5.3722
= 7/5.3722= 1.303
= P ( Z <1.303) From Standard NOrmal Table
= 0.9037
P(X > = 118) = (1 - P(X < 118))
= 1 - 0.9037 = 0.0963
c.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 111) = (111-111)/5.3722
= 0/5.3722 = 0
= P ( Z <0) From Standard Normal Table
= 0.5
P(X < 118) = (118-111)/5.3722
= 7/5.3722 = 1.303
= P ( Z <1.303) From Standard Normal Table
= 0.90371
P(111 < X < 118) = 0.90371-0.5 = 0.4037
d.
P(X < 96) = (96-111)/5.3722
= -15/5.3722= -2.7922
= P ( Z <-2.7922) From Standard NOrmal Table
= 0.0026
option C is correct

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