A 2014 Pew study found that the average US Facebook user has 338 friends. The st

Question

A 2014 Pew study found that the average US Facebook user has 338 friends. The study also found that the median US Facebook user has 200 friends. What does this imply about the distribution of the variable "number of Facebook friends"?

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 102 Facebook users, what is the probability that the mean number of friends will be less than 343? _______

b. If we randomly sample 119 Facebook users, what is the probability that the mean number of friends will be less than 318? _______

c. If we randomly sample 500 Facebook users, what is the probability that the mean number of friends will be greater than 343? _______
(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 _____ and _____

e. The 75th percentile of the sampling distribution of mean number of friends, from samples of size n=102, is: ______

THANKS!

Explanation / Answer

As median lies to the left of the mean on histogram or data curve, this indicates the data is more in the right tail of the curve. Hence data is right skewed.

mean = 338
std. dev. = 194

a)
SE = 194/sqrt(102) = 19.2089

P(X < 343)
= P(z < (343 - 338)/19.2089)
= P(z < 0.2603)
= 0.6027 (use standard normal z table to find this value)

b)
SE = 194/sqrt(119) = 17.784

P(X < 318)
= P(z < (318 - 338)/17.784)
= P(z < -1.1246)
= 0.1304 (use standard normal z table to find this value)

c)
SE = 194/sqrt(500) = 8.6759

P(X > 343)
= P(z > (343 - 338)/8.6759)
= P(z > 0.5763)
= 0.2822 (use standard normal z table to find this value)

d)

e)

SE = 194/sqrt(102) = 19.2089

z-value for 75 percentile using standard normal table is 0.6745

xbar = mean + z*SE
xbar = 338 + 0.6745*19.2089
xbar = 350.9564

Hence 75 percentile is 351

CI for 95% n 500 mean 338 z-value of 95% CI 1.9600 std. dev. 194 SE = std.dev./sqrt(n) 8.67594 ME = z*SE 17.00454 Lower Limit = Mean - ME 320.99546 Upper Limit = Mean + ME 355.00454 95% CI (320.9955 , 355.0045 )

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