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"? (You have two attempts for this problem, and five attempts each for the remaining problems) The distribution is normal The distribution is bimodal The distribution is approximately Q3 The distribution is left skewed The distribution is trending The distribution is right skewed The distribution is symmetrical Correct: Your answer is correct. 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 191. Let's also suppose that 338 and 191 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 116 Facebook users, what is the probability that the mean number of friends will be less than 346? 0.6736 Correct: Your answer is correct. b. If we randomly sample 104 Facebook users, what is the probability that the mean number of friends will be less than 316? 0.121 Correct: Your answer is correct. c. If we randomly sample 400 Facebook users, what is the probability that the mean number of friends will be greater than 346? 0.2033 Incorrect: Your answer is incorrect. (Round to the nearest integer for parts d. and e.) d. If we repeatedly take samples of n=400 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=116, is:

I couldn't get C and kept trying, I can only calculate D once I answer C correctly so can't calculate D or E

Answer 1

c)

for normal distribution z score =(X-μ)/σ
here mean=       μ=	338
std deviation   =σ=	191.000
sample size       =n=	400
std error=σx̅=σ/√n=	9.5500

probability that the mean number of friends will be greater than 346:

probability =

P(X>346)

=

P(Z>0.84)=

1-P(Z<0.84)=

1-0.7995=

0.2005

d)

for 95% middle values ; z =-/+1.96

hence corresponding value =mean-/+ z*Std deviation=338-/+1.96*9.55 =319.28 to 356.72

e)

for 75th percentile ; z=0.67

corresponding value =mean+ z*Std deviation=338+0.67*17.7339 =349.88