An exhaustive study of all active Facebook accounts was recently conducted by Fa

Question

An exhaustive study of all active Facebook accounts was recently conducted by Facebook. One variable Facebook recorded was the number of friends X of each Facebook user. Suppose X has expected value E(X) = 184 and standard deviation SD(X) = 281.5. Since the possible values of X are only integers and since the distribution of X is highly skewed to the right, the distribution of X cannot be described by a normal model. Suppose you select a random sample of 33 Facebook users and record the number of Facebook friends each user has.

Question 1. What is the probability that the 33 Facebook users in your sample have a sample mean number of friends x greater than 175?

(use 4 decimal places in your answer)

Question 2. What is the probability that the 33 Facebook users in your sample have a sample mean number of friends x less than 184?

Explanation / Answer

An exhaustive study of all active Facebook accounts was recently conducted by Facebook. One variable Facebook recorded was the number of friends X of each Facebook user. Suppose X has expected value E(X) = 184 and standard deviation SD(X) = 281.5. Since the possible values of X are only integers and since the distribution of X is highly skewed to the right, the distribution of X cannot be described by a normal model. Suppose you select a random sample of 33 Facebook users and record the number of Facebook friends each user has.

Since sample size 33 is greater than 30, sample mean follows approximately normal, whatever the population distribution.

Question 1. What is the probability that the 33 Facebook users in your sample have a sample mean number of friends x greater than 175?

(use 4 decimal places in your answer)

Standard error =sd /sqrt(n) =281.5/sqrt(33) =49.0029

Z value for 175, z = (175-184) /49.0029 = -0.18

P( mean x >175) = p( z > -0.18) =0.5714

Question 2. What is the probability that the 33 Facebook users in your sample have a sample mean number of friends x less than 184?

Z value for 184, z = (184-184) /49.0029 = 0

P( mean x < 184) = p( z < 0) =0.5

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