A social media survey found that 68 of parents are friends w

A social media survey found that 68% of parents are \"friends\" with their children on a certain online networking site. A random sample of 140 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 99 or more parents from this sample are \"friends\" with their children on this online networking site? P(99 or more parents from this sample are \"friends\" with their children) = (   ).

(Round to four decimal places as needed.)

c. What is the probability that between 95 and 104 parents from this sample are \"friends\" with their children on this online networking site? P(Between 95 and 104 parents from this sample are \"friends\" with their

children) = (   ).

(Round to four decimal places as needed.)

d. If 81 parents responded that they are \"friends\" with their children on this online networking site, does this result support the findings of the social media survey?

A. Yes, the probability of obtaining a result this small or smaller given that the findings of the social media survey are accurrate is less than 0.05.

B. Yes, the probability of obtaining a result this small or smaller given that the findings of the social media survey are accurrate is greater than 0.05.

C. No, the probability of obtaining a result this small or smaller given that the findings of the social media survey are accurrate is greater than 0.05.

D. No, the probability of obtaining a result this small or smaller given that the findings of the social media survey are accurrate is less than 0.05.

Solution

Mean ( np ) = 95.2
Standard Deviation ( npq )= 140*0.68*0.32 = 5.5194
Normal Distribution = Z= X- u / sd                   
a)
Proportion ( P ) =0.68
Standard Error = Sqrt(p*(1-p)/n))
x = Mean
n = Sample Size,
Standard Error = Sqrt ( (0.68*0.32) /140) )
= 0.0394
                  

b)
P(X < 99) = (99-95.2)/5.5194
= 3.8/5.5194= 0.6885
= P ( Z <0.6885) From Standard Normal Table
= 0.7544                  
P(X > = 99) = (1 - P(X < 99)
= 1 - 0.7544 = 0.2456                  
c)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 95) = (95-95.2)/5.5194
= -0.2/5.5194 = -0.0362
= P ( Z <-0.0362) From Standard Normal Table
= 0.48555
P(X < 104) = (104-95.2)/5.5194
= 8.8/5.5194 = 1.5944
= P ( Z <1.5944) From Standard Normal Table
= 0.94457
P(95 < X < 104) = 0.94457-0.48555 = 0.459