For part a I have tried (.3745,.6255,.1.3746) and neither of those were right..
ID: 3073518 • Letter: F
Question
For part a I have tried (.3745,.6255,.1.3746) and neither of those were right.. not too sure what I did wrong compared to what I did on b..(I love to rate :) 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, whr 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 symmetrical -" The distribution is trending The distribution is normal The distribution is right skewed The distribution is bimodal The distribution is approximately Q3 The distribution is left skewed The Pew study did not report a standard deviation, but given the number of Facebook friends is highly veriable, let's suppose that the standard deviation is 205. Cet's also suppose that 338 and 205 are population values (they aren't, but we don't know the true popuiation 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 117 Facebook users, what is the probability that the mean number of friends will be less than 344 b. If we randomly sample 115 Facebook users, what is the probability that the mean number of friends will be less than 3127 c869 c. wo randomly senle 400 Facebook users, what is the probability that the meen number of friends wa be yeater than 3447 Round to the nearest integer for parts d. and e.) d. If we repeatedly take samples o na400 Facebook users and construct a sam pling distribution of mean number of friends, we should expect that 95% o.. ple means wil le between and e. The 75th percentile of the sampling distribution of mean number of friends, from samples of size n 117,is:Explanation / Answer
a)
mean = 338, sd = 205, n = 117
SE = 205/sqrt(117) = 18.9523
P(X < 344)
= P(z < (344 - 338)/18.9523)
= P(z < 0.3166)
= 0.624
c)
mean = 338, sd = 205, n = 400
SE = 205/sqrt(400) = 10.2500
P(X > 344)
= P(z > (344 - 338)/10.2500)
= P(z > 0.5854)
= 0.279
d)
ME = z*SE = 1.96*10.25 = 20.0896
Lower Limit = Mean - ME = 338 - 20.089 = 317.91037
Upper Limit = Mean + ME = 338 + 20.0896 = 358.08963
95% CI (318 , 358)
e)
SE = 205/sqrt(117) = 18.9523
z-value = 0.6745
x = 338 + 0.6745*18.9523
x = 350.7833264
Ans: 351
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