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2 July, 06:07

Early in August an undergraduate college discovers that it can accommodate a few extra students. Enrolling these additional students would provide a substantial increase in revenue without increasing the operating costs of the college; that is, no new classes would have to be added. From past experience, the college knows that 40% of those students will actually enroll. a) What is the probability that at most six students will enroll if the college offers admission to ten more students? b) What is the probability that more than 12 will actually enroll if admission is offered to 20 students? c) If 70% of those students admitted actually enroll, what is the probability that at least 12 out of 15 students will actually enroll?

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  1. 2 July, 07:07
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    A. P (x ≤ 6) = 0.9452

    B. P (x > 12) = 0.021

    C. P (x ≥ 12) = 0.2968

    Step-by-step explanation:

    Let's recall that in a binomial distribution the probability of success p in each trial is a fixed value and the result of each trial is independent of any previous trial.

    A. Let's find out the probability that at most six students will enroll if the college offers admission to ten more students, using the following binomial distribution table:

    Binomial distribution (n=10, p=0.4)

    f (x) F (x) 1 - F (x)

    x Pr[X = x] Pr[X ≤ x]

    0 0.0060 0.0060

    1 0.0403 0.0464

    2 0.1209 0.1673

    3 0.2150 0.3823

    4 0.2508 0.6331

    5 0.2007 0.8338

    6 0.1115 0.9452

    7 0.0425 0.9877

    8 0.0106 0.9983

    9 0.0016 0.9999

    10 0.0001 1.0000

    P (x ≤ 6) = 0.9452

    B. Let's find that more than 12 will actually enroll if admission is offered to 20 students, using this second binomial distribution table:

    Binomial distribution (n=20, p=0.4)

    f (x) F (x) 1 - F (x)

    x Pr[X = x] Pr[X ≤ x]

    0 0.0000 0.0000

    1 0.0005 0.0005

    2 0.0031 0.0036

    3 0.0123 0.0160

    4 0.0350 0.0510

    5 0.0746 0.1256

    6 0.1244 0.2500

    7 0.1659 0.4159

    8 0.1797 0.5956

    9 0.1597 0.7553

    10 0.1171 0.8725

    11 0.0710 0.9435

    12 0.0355 0.9790

    13 0.0146 0.9935

    14 0.0049 0.9984

    15 0.0013 0.9997

    16 0.0003 1.0000

    17 0.0000 1.0000

    18 0.0000 1.0000

    19 0.0000 1.0000

    20 0.0000 1.0000

    P (x ≤ 12) = 0.979, then:

    P (x > 12) = 1 - 0.979 = 0.021

    C. Let's find the probability that at least 12 out of 15 students will actually enroll if 70% of those students admitted actually enroll, using this new binomial distribution table:

    Binomial distribution (n=15, p=0.7)

    f (x) F (x) 1 - F (x)

    x Pr[X = x] Pr[X ≤ x]

    0 0.0000 0.0000

    1 0.0000 0.0000

    2 0.0000 0.0000

    3 0.0001 0.0001

    4 0.0006 0.0007

    5 0.0030 0.0037

    6 0.0116 0.0152

    7 0.0348 0.0500

    8 0.0811 0.1311

    9 0.1472 0.2784

    10 0.2061 0.4845

    11 0.2186 0.7031

    12 0.1700 0.8732

    13 0.0916 0.9647

    14 0.0305 0.9953

    15 0.0047 1.0000

    P (x ≤ 11) = 0.7031, then:

    P (x ≥ 12) = 1 - 0.7031 = 0.2969, or,

    P (x ≥ 12) = P (12) + P (13) + P (14) + P (15)

    P (x ≥ 12) = 0.17 + 0.0916 + 0.0305 + 0.0047

    P (x ≥ 12) = 0.2968
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