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"? (You have two attempts for this problem, and five attempts each for the remaining problems) The distribution is right skewed The distribution is bimodal The distribution is normal The distribution is left skewed The distribution is trending The distribution is approximately Q3 The distribution is symmetrical 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 193. Let's also suppose that 338 and 193 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 119 Facebook users, what is the probability that the mean number of friends will be less than 343? b. If we randomly sample 104 Facebook users, what is the probability that the mean number of friends will be less than 320? 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=119, is

Explanation / Answer

1)as mean is higher then median therefore the distribution is right skewed

2)

a)for n=119 ; std error of mean =std deviation/(n)1/2 =17.69

therefore P(X<343)=P(Z<(343-338)/17.69)=P(Z<0.2826)=0.6113

b)

for n=104 ; std error of mean =std deviation/(n)1/2 =18.93

therefore P(X<343)=P(Z<(320-338)/17.69)=P(Z<-0.9511)=0.1708

c)

for n=500 ; std error of mean =std deviation/(n)1/2 =8.63

therefore P(X>343)=P(Z>(343-338)/8.63)=P(Z>0.5793)=0.2812

d) for 95% CI ; z=1.96

therefore lie between mean -./+ z*std error =321 and 355

e) for 75 th percentile z score =0.6745

corresponding score =mean +z*std error =350

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