Conditions of 10% and random are met, probably (unclear)
4.7% chance
0.95^3 = 86.3%
N(10.2, 0.12/sqrt(3))
0.18% chance
near 0
we cant apply the normal model to the individual since the model wwould not fit a skwed distribution. Therefore our answer would be far away from the empiracal actual value
We cannot know, since we havent yet learned how to deterime whether quanitivative values are enough to representative
Probably, if this is a representative number of people and fits out conditions for using the normal model through the central limit thorem.
N(9984, 353)
0.48
$ 9531
30.8 % chance of more than 50 pounds
sd = 7.8
69.6 %
0.037
0.325 chance for 10 asheyers to average 5 pounds higher