MAS 183 Assignment 2 PDF

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MAS 183 Assignment 2 1. a)

HIV Test Results

Patient with HIV? Yes

No

Total

Positive

0.1176

0.0352

0.1528

Negative

0.0024

0.8448

0.8472

Total

0.12

0.88

1

b)

P(Has HIV | HIV positive) = 0.1176/0.1528 = 0.7696 = 76.98%

c)

P(No HIV | HIV negative) = 0.8448/0.8472 = 0.9972 = 99.72%

d)

According to the statement given, the clinic uses a diagnostic test which returns a

positive result in 98% of cases wherein the patient actually has HIV, meanwhile 96% to those who don’t, so the diagnostic test does better in detecting those with HIV.

2. For X to be considered an observation from a binomial distribution, it has to have all the following conditions: a) Definite number of trials (n) b) Trials are independent to each other c) Only two possible outcomes from each trial (success or fail) d) The same probability (p) of success for each trial e) The variable X is the number of successes out of ‘n’ trials The test described does not meet the criteria mentioned above as the number of trials must be fixed. The result of the test described is the pedestrian passing in or out during a 15-minute period, however, the total number of trials (pedestrian entry) are not fixed. Moreover, this trial is trying to predict the probability of certain events, based on

continuous events, therefore, this trial follows a Poisson distribution and not binomial distribution.

3. a) Bin(n,p) Where, n is the number of trials = 25 p = 0.3 Thus, the probability distribution of X: X ~ Bin (25, 0.3)

b) � , = np = 25 x 0.3 = 7.5

i.

√ np ( 1− p)

σ 

=

√ 25 x 0.3 x 0.7

= 2.29

P(X > 9) = 1 – P(x  9) ii.

= 1 - (P(X = 0) + P(X = 1) + P(X = 2) + p(X = 3) + P(X = 4) + P(X  = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9)) = 1 – 0.8106 = 0.1894 iii.

P(5  x  11)



 P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) +

P(X = 10) + P(X = 11) = 0.8653 

c) P(X = 2) = 25C2 x (0.3)2 x (0.7)23 = 0.0074 Since the probability is less than 0.05, so it is unusual.

4.

a) z =

265 −272.3 8.8

= -0.8295

z=

279 −272.3 8.8

= 0.7614

= P( z < 0.7614) – P( z < -0.8295) = P(z < 0.7614) – ((1 – P( z < 0.8295) = 0.7768 – (1-0.7966) = 0.5734

b) z =

288−272. 3 8.8

= 1.7841

P(x > 288) = P(z > 1.7841) = 1 – 0.9628 = 0.0372 c) First quartile = 25th percentile = � + z x σ, where z is critical value at 25% confidence level = 272.3 + (-0.6745) x 8.8 = 266.3644 Third quartile = 75th percentile = � + z x σ, where z is critical value at 75% confidence level = 272.3 + (0.6745) x 8.8 = 278.2356

5. a)

This method used to collect data for this survey can be considered as convenience sampling. The data was easily obtained, and potentially bias. Not only that, but it would be wrong to interpret results from the method used as public opinion. The number of people giving their opinion is limited and would not even be close to half of the population, thus making the results not useful. Furthermore, it would be wrong for the TV station to interpret the results of the survey because if it is not majority of the population that participated, then it should not be considered as public opinion.

b) i.

As MAS 183 is a prerequisite unit for science students, they may be used to represent Murdoch University in relation to attitudes to public policy on climate change, but the question lies on whether they are very well concerned, knowledgeable, and aware on climate change. However, if being compared to other students not taking MAS 183, then MAS 183 students may be the best option as representatives of Murdoch University.

ii.

The brand of mobile phones has no correlation to a specific degree or unit. So, whether MAS 183 students are likely to be representatives of Murdoch University students in relation to their brand of mobile phone does not relate to anything and the possibility of others not owning a mobile phone is also another thing to consider....