Can someone please help me out with this problem and explain how they got the an

Question

Can someone please help me out with this problem and explain how they got the answers? I have no idea how to do it or where to start

aoo State UniLogin - Canvas | Colc PDF to PPT Convert WebAssign Stats Book Other bool A 2014 Pew study found that the avera distribution of the variable "number of Facebook friends"? (you have two attempts for this problem, and five attempts each for the remaining problems) nas 338 friend that the median US Facebook user has 200 friends. What does this imply about the ® The distribution is left skewed O The distribution is right skewed ® The distribution is symmetrical The distribution is trending The distribution is bimodal The distribution is normal The distribution is approximately Q3 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 deviat 338 and 207 are population values (they aren't, but we don't know the true population values so this is the best we can doj. (Use 3 decimal place precision fo suppose that b., and c.) r part a. If we randomly sample 114 Facebook users, what is the probability that the mean number of friends will be less than 347? If we randomly sample 100 Facebook users, what is the probability that the mean number of friends will be less than 318? e randomly sample 400 Facebook users, what is the probability that the mean number &f friends will be greater than 3472 to the nearest integer for parts d and e. ng eistribu on sean number of friends, we should expect that 95% of sample means will d. If we repeatedly take samples of n-400 Facebook users and construct asam and e. The 75th percentile of the sampling distribution of mean number of friends, from samples of size n#114, is: 1

Explanation / Answer

as mean is higher than median therefore

the distribution is right skewed.

a)

( please try 0.679 ifg this comes wrng due to rounding)

b)

( please try 0.167 if above comes wrng)

d)

d)

for middle 95%; critical z -=1.96

hence correspinding values =mean-/+z*Std deviation =338-/+ 1.96*10.35 =318 to 358

e)

for 75th percentile critical value of z =0.67

hence correspinding values =mean+z*Std deviation =338+0.67*19.3873 =351

for normal distribution z score =(X-)/ here mean=       = 338 std deviation   == 207.000 sample size       =n= 114 std error=x=/n= 19.3873

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