GEM2900, 2013/2014 Semester II Assignment 02

Answers

1. The answer is (c).

Number of ways to choose 2 days out of 7 days:

(7

2

).

Number of ways to choose 2 weekdays out of 5:

(5

2

).

Required probability is then (5

2

)/

(7

2

)=

10

21 0.476.

2. The answer is (c).

For the first blouse, there are six blouses of which one is red, so the chance of selecting ared blouse is 1/6.

For the second blouse, we condition on whether this first blouse chosen is red or not. LetRi denote the event that the ith blouse is red.

P (R2) = P (R2|R1)P (R1) + P (R2|Rc1)P (Rc1) = 0 1/6 + 1/5 5/6 = 1/6.So the two probabilities are the same.

3. The answer is (d).

Let A be the event that the wallet originally contained a $5 note. Let B be the eventthat the note removed is a $5 note. The desired probability is given by

P (A|B) = P (B|A)P (A)P (B|A)P (A) + P (B|Ac)P (Ac) =

1 1/21 1/2 + 1/2 1/2 = 2/3 0.67.

4. (a) Inductive reasoning. The experiment (mice data) leads to a hypothesis (inhibition oftumour growth).

(b) Inductive reasoning. The experiments (leukaemia cells) lead to a hypothesis (certainsubstances induce programmed cell death).

(c) Deductive reasoning. The hypothesis (DLC1 gets reactivated) makes a clear predic-tion (different tumour cell lines should be affected differently by the herb).

5. The answer is (b).

Note that X N(34, 12). HenceP [4X 2 110] = P [4X 112]

= P [X 28]= P [N(34, 12) 28] = P [N(0, 12) 6]= P [N(0, 1) 6/12] = (0.5) 0.3085.

(1)

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GEM2900, 2013/2014 Semester II Assignment 02

6. The sample mean is 156, the sample standard deviation is 28.5.

The standard error is 6.91 (rounded to three significant digits, e.g. 4.32).

Our rule-of-thumb interval around the sample mean is given by [ 145.6 , 166.4 ] (roundedto four significant digits, e.g. 143.2)

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