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 200. Let's also suppose that 338 and 200 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 117 Facebook users, what is the probability that the mean number of friends will be less than 347?  

b. If we randomly sample 106 Facebook users, what is the probability that the mean number of friends will be less than 316?  

c. If we randomly sample 600 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=600 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=117, is:

Answer 1

In this case, Mean = 338 ; Median = 200

So, Mean is greater than Median in this case. This tells us that the distribution is Right - Skewed. It has a long tail on its right side.

A. Sample Size, n = 117

Mean = 338 ; Standard deviation = 200

Sample Standard Deviation,s = 200 /   = 200/ = 18.49

Now, Standard Normal Probability Distribution , Z = (X - Mean ) / s

Probability that the mean number of friends will be less than 347, P(X<347) = P(Z < (X - Mean ) / s) = P(Z < (347 - 338)/18.49) = P( Z < 0.4865)

Now, from Z table , P(Z < 0.4865) = 0.687.

Therefore, P(X<347) = 0.687 = 68.7%

B.

Sample Size, n = 106

Mean = 338 ; Standard deviation = 200

Sample Standard Deviation,s = 200 /   = 200/ = 19.425

Now, Standard Normal Probability Distribution , Z = (X - Mean ) / s

Probability that the mean number of friends will be less than 316, P(X<316) = P(Z < (X - Mean ) / s)

= P(Z < (316 - 338)/19.425) = P( Z < -1.1325)

Now, from Z table , P(Z < 0.4865) = 0.0926 = 9.26%

Therefore, P(X<316) = 0.0926 = 9.26%

C.

Sample Size, n = 600

Mean = 338 ; Standard deviation = 200

Sample Standard Deviation,s = 200 /   = 200/ = 8.165

Now, Standard Normal Probability Distribution , Z = (X - Mean ) / s

Probability that the mean number of friends will be less than 347, P(X<347) = P(Z < (X - Mean ) / s)

= P(Z < (347 - 338)/8.165) = P( Z < 1.1023)

Now, from Z table , P(Z < 1.1023) = 0.8686 = 86.86%

Therefore, P(X<347) = 0.8686 = 86.86%

D. If we repeatedly take samples of 600 and construct a sampling distribution, then the distribution will be same as the Population.

Now, Mean = 338 ; Standard deviation = 200

Therefore 95% CI is given as, CI = ( Mean - 1.96* Standard Deviation, Mean + 1.96* Standard Deviation)

= ( 338 - 1.96* 200, 338 + 1.96* 200) = (-54, 730)

Therefore 95% CI = (0,730).

Hope I answered your query.