Suppose that the probability of getting an A in a particu- lar course is 0.15, a
ID: 3124831 • Letter: S
Question
Suppose that the probability of getting an A in a particu- lar course is 0.15, and assume that the all student grades are independent. If you randomly sample 15 students taking the course;
(a) Find the expected number of students out of the 15 that will get an A, and the standard deviation of number of students out of the 15 that will get an A.
(b) Find the probability that no student gets an A.
(c) Find the probability that at most 2 students get an A.
(d) Find the probability that more than the expected number of students get an A.
Explanation / Answer
a)
As n = 15, p = 0.15,
E(x) = n p = 15*0.15 = 2.25 [ANSWER]
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b)
Note that the probability of x successes out of n trials is
P(n, x) = nCx p^x (1 - p)^(n - x)
where
n = number of trials = 15
p = the probability of a success = 0.15
x = the number of successes = 0
Thus, the probability is
P ( 0 ) = 0.087354219 [ANSWER]
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c)
Using a cumulative binomial distribution table or technology, matching
n = number of trials = 15
p = the probability of a success = 0.15
x = the maximum number of successes = 2
Then the cumulative probability is
P(at most 2 ) = 0.604225204 [ANSWER]
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d)
That means at least 3 gets an A.
Note that P(at least x) = 1 - P(at most x - 1).
Using a cumulative binomial distribution table or technology, matching
n = number of trials = 15
p = the probability of a success = 0.15
x = our critical value of successes = 3
Then the cumulative probability of P(at most x - 1) from a table/technology is
P(at most 2 ) = 0.604225204
Thus, the probability of at least 3 successes is
P(at least 3 ) = 0.395774796 [ANSWER]
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