The Pew study did not report a standard deviation, but given the number of Faceb

Question

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 205. Let's also suppose that 338 and 205 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 120 Facebook users, what is the probability that the mean number of friends will be less than 341? b. If we randomly sample 117 Facebook users, what is the probability that the mean number of friends will be less than 315? c. If we randomly sample 500 Facebook users, what is the probability that the mean number of friends will be greater than 341? 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=120, is:

Explanation / Answer

a)

b)

c)

d)

for 95% CI ; z= -/+ 1.96

hence 95% CI =mean -/+z*std deviation =482 ; 518

e)

for 75th percentile z =0.6745

corresponding value =338+0.6745*18.7139 =351

for normal distribution z score =(X-)/ here mean=       = 338.000 std deviation   == 205.00 sample size       =n= 120 std error=x=/n= 18.7139

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