I can\'t figure out A. and E. ttpsJ/www O The distribution is left skewed The Pe

Question

I can't figure out A. and E.

ttpsJ/www O The distribution is left skewed 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 210. Let also suppose that 338 and 210 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 346? b. If we randomly sample 119 Facebook users, what is the probability that the mean number of friends will be less than 310? 0.0735 c. If we randomly sample 600 Facebook users, what is the probability that the mean number of friends will be greater than 346? 0.176 Round to the nearest integer for parts d. and e.) d. If we repeatedly take samples o n-600 Facebook users and construct a sampling di tribution or mean num means will lie between 321 rofmeds ve hould expect at 5% a pie and 355 e. The 75th percentile of the sampling distribution of mean number of friends, from samples Submit Answer Save Progress Practice Another Version 46 PM ^ Type here to search

Explanation / Answer

Answer:

a).

standard error = 210/sqrt(119) = 19.2507

z value for 346, z =(346-338)/19.2507 = 0.42

P( mean x <346) = P( z < 0.42) =0.6628

=0.663 ( rounded to 3 decimals)

e).

z value for 75th percentile = 0.674

x = mean+z*sd = 338+0.674*19.2507 =350.975

The required x= 351 ( rounded to whole number)

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