Question

The Pew study did not report a standard deviation, but given the number of Facebook friends is highly variable, let's suppose that the standard deviation is 199. Let's also suppose that 338 and 199 are population values (they aren't, but we don't know the true population values so this is the best we can do). (Use 3 decimal place precision for parts a., b., and c.)

a. If we randomly sample 108 Facebook users, what is the probability that the mean number of friends will be less than 345?  

b. If we randomly sample 103 Facebook users, what is the probability that the mean number of friends will be less than 311?  

c. If we randomly sample 500 Facebook users, what is the probability that the mean number of friends will be greater than 345?

d. If we repeatedly take samples of n=500 Facebook users and construct a sampling distribution of mean number of friends, we should expect that 95% of sample means will lie between  and  

e. The 75th percentile of the sampling distribution of mean number of friends, from samples of size n=108, is:

Answer 1

Solution:- Given that mean = 338, sd = 199

a. n = 108 , X = 345

P(X < 345) = P((X - μ)/(σ/sqrt(n)) <(345-338)/(199/sqrt(108))

= P(Z < 0.3656)

= 0.6443

= 0.644(rounded)

b. n = 103, X = 311

P(X < 311) = P((X - μ)/(σ/sqrt(n)) < (311-338)/(199/sqrt(103))

= P(Z < -1.3770)

=  0.0838

= 0.084(rounded)

c. n = 500, X = 345

P(X > 345) = P((X - μ)/(σ/sqrt(n)) > (345-338)/(199/sqrt(500))

= P(Z > 0.7866)

= 0.2148

= 0.215(rounded)

d. 95% confidence interval for the mean = X +/- Z*s/sqrt(n)

= 338 +/- 1.96*199/sqrt(500)

= 320.5568 , 355.4431

e. 75th percentile , Z = z=0.6745

x = 338 + 0.6745*199/sqrt(500) = 344.0027