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?

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)

xPr[X = x]Pr[X ≤ x]

00.00600.0060

10.04030.0464

20.12090.1673

30.21500.3823

40.25080.6331

50.20070.8338

60.11150.9452

70.04250.9877

80.01060.9983

90.00160.9999

100.00011.0000

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)

xPr[X = x]Pr[X ≤ x]

00.00000.0000

10.00050.0005

20.00310.0036

30.01230.0160

40.03500.0510

50.07460.1256

60.12440.2500

70.16590.4159

80.17970.5956

90.15970.7553

100.11710.8725

110.07100.9435

120.03550.9790

130.01460.9935

140.00490.9984

150.00130.9997

160.00031.0000

170.00001.0000

180.00001.0000

190.00001.0000

200.00001.0000

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)

xPr[X = x]Pr[X ≤ x]

00.00000.0000

10.00000.0000  

20.00000.0000

30.00010.0001

40.00060.0007

50.00300.0037

60.01160.0152

70.03480.0500

80.08110.1311

90.14720.2784

100.20610.4845

110.21860.7031

120.17000.8732

130.09160.9647

140.03050.9953

150.00471.0000

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

a .4y4

hope this !

14.3

step-by-step explanation:

add all the numbers together.

9 + 12 + 15 + 16 + 17 + 17 = 86

now, divide 86 by how many numbers we have: 6.

86 / 6 = 14.3