Question 972837: A local band is going on a U.S. Summer tour, and they average about 2000 people per concert, with a standard deviation of about 400. Assume that these concert numbers follow a normal distribution.
a) if a concert is selected at random, what is the probability that there were more than 2500 people at the concert?
b) if a concert is selected at random, what is the probability that there were less than 1800 people at the concert?
c) if a concert is selected at random, what if the probability that there were between 1800 and 2500 people at that concert?
d) for a concert to be in the top 10% as far as attendance, at least how many people would need to attend the concert?
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! mean=400, sd=400
z= (x-mean)/sd
a) (2500-2000)/400
z=500/400, or +1.25
Probability more than that is 0.106
b) (1800-2000)/400
z=-0.5, and want probability to the left
P=0.309
c) From the first two, we have the probability of what we don't want. Therefore, the probability of what we DO want is the complement of that or 1- the sum of the first two, which add to 0.415. Therefore, the desired probability is (1-0.415)=0.585
d)The 90th percentile is z=1.282
1.282= (x-2000)/400
Multiply both sides by 400
512.62=(x-2000)
x=2513 people, rounding up, needing an integer for the answer.
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