Question 1178780: Americans receive an average of 18 Christmas cards each year. Suppose the number of Christmas cards is normally distributed with a standard deviation of 5. Let X be the number of Christmas cards received by a randomly selected American. Round all answers to 4 decimal places where possible.
a. What is the distribution of X? X ~ N( ,)
b. If an American is randomly chosen, find the probability that this American will receive no more than 21 Christmas cards this year.
(Round z-score up to 2 decimal places.)
c. If an American is randomly chosen, find the probability that this American will receive between 19 and 24 Christmas cards this year.
(Round z-score up to 2 decimal places.)
d. 82% of all Americans receive at most how many Christmas cards? (Please enter a whole number)
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! X~N(18, 25) mean, sigma^2
z <(21-18)/5=3/5. That probability is 0.7257
z will be between 1/5 and 6/5. That probability is 0.3057
82% is a z of 0.9154
0.9154=(x-18)/5
x=18+4.58
22.58 cards.
At most it would be 22 cards.
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