The Pew study did not report a standard deviation, but given the number of Faceb

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 194. Let's also suppose that 338 and 194 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 112 Facebook users, what is the probability that the mean number of friends will be less than 347? b. If we randomly sample 111 Facebook users, what is the probability that the mean number of friends will be less than 315? c. If we randomly sample 500 Facebook users, what is the probability that the mean number of friends will be greater than 347? (Round to the nearest integer for parts d. and e.) 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 e. The 75th percentile of the sampling distribution of mean number of friends, from samples of size n=112, is and

Explanation / Answer

mean = 338
std. dev. = 194

a)
P(X < 347) = P(z < (347 - 338)/(194/sqrt(112))) = P(z < 0.491) = 0.6883

b)
P(X < 315) = P(z < (315 - 338)/(194/sqrt(111))) = P(z < -1.2491) = 0.1058

c)
P(X > 347) = P(z > (347 - 338)/(194/sqrt(500))) = P(z < 1.0374) = 0.1498

d)
Margin of error, ME = z*sigma/sqrt(n)
z = 1.96 for 95% CI
ME = 1.96*194/sqrt(500) = 17.0048

CI = (mean - ME, mean + ME) = (338 - 17, 338 + 17) = (321, 355)

(e)
For 75th percentile, z = 0.6745
x = mean + z*sigma/sqrt(n) = 338 + 0.6745*194/sqrt(112) = 350.3644 i.e. 350

Hence 75th percentile is 350

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