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(i) An experiment was conducted to study the choices made in fund selection. Su...

May 17, 2024

Solution by Steps

step 1

Calculate the total number of students:

100 + 100 = 200

step 2

Calculate the number of students who selected the lowest-cost fund:

14 + 14 = 28

step 3

Calculate the number of students who selected the second-lowest-cost fund:

40 + 35 = 75

step 4

Calculate the total number of students who selected the lowest- or second-lowest-cost fund:

28 + 75 = 103

step 5

Calculate the probability:

\frac{103}{200}

Answer

$\frac{103}{200}$

# b. Probability of selecting the lowest-cost fund and being an undergraduate

step 1

Calculate the number of undergraduate students who selected the lowest-cost fund:

14

step 2

Calculate the total number of students:

200

step 3

Calculate the probability:

\frac{14}{200}

Answer

$\frac{14}{200}$

# c. Probability of selecting the highest-cost fund, given the student is an undergraduate

step 1

Calculate the number of undergraduate students who selected the highest-cost fund:

27

step 2

Calculate the total number of undergraduate students:

100

step 3

Calculate the conditional probability:

\frac{27}{100}

Answer

$\frac{27}{100}$

# d. Probability of being an MBA student, given the student selected the lowest-cost fund

step 1

Calculate the number of MBA students who selected the lowest-cost fund:

14

step 2

Calculate the total number of students who selected the lowest-cost fund:

28

step 3

Calculate the conditional probability:

\frac{14}{28}

Answer

$\frac{14}{28}$

Part (ii) # a. Probability that no private airplane will arrive in a given hour

step 1

Use the Poisson distribution formula:

P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}

step 2

Set

\lambda = 1.4

and

k = 0

P(X = 0) = \frac{1.4^0 e^{-1.4}}{0!}

step 3

Simplify:

P(X = 0) = e^{-1.4}

Answer

$e^{-1.4}$

# b. Probability that less than 5 private airplanes will arrive in an 8-hour interval

step 1

Calculate the new

\lambda

for 8 hours:

\lambda = 1.4 \times 8 = 11.2

step 2

Use the cumulative Poisson distribution: P(X < 5) = \sum_{k=0}^{4} \frac{11.2^k e^{-11.2}}{k!}

step 3

Calculate each term and sum: P(X < 5) = \frac{11.2^0 e^{-11.2}}{0!} + \frac{11.2^1 e^{-11.2}}{1!} + \frac{11.2^2 e^{-11.2}}{2!} + \frac{11.2^3 e^{-11.2}}{3!} + \frac{11.2^4 e^{-11.2}}{4!}

Answer

$\sum_{k=0}^{4} \frac{11.2^k e^{-11.2}}{k!}$

Key Concept

Probability calculation using Poisson distribution

Explanation

The Poisson distribution is used to model the number of events occurring within a fixed interval of time or space.