Term 1 Assessment 1 Solutions (STAT0021) PDF

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Term 1 Assessment 1: SolutionsThomas HonnorLast updated: 29 November, 2020Note that Assessment 1 was setup as a Moodle quiz, in such a way that some of the numbers involved in each of the questions varied between different students. It isn’t really feasible for me to include in these solutions to ...

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Term 1 Assessment 1: Solutions Thomas Honnor Last updated: 29 November, 2020 Note that Assessment 1 was setup as a Moodle quiz, in such a way that some of the numbers involved in each of the questions varied between different students. It isn’t really feasible for me to include in these solutions to the numerical results for every one of the different variants. Instead, I’ll provide the solution methodolgy and the numerical results for one variant, from which you can follow through the process to see how the final answer was obtained for your variants.

1. . Events

and

Calculate

. are independent. .

You should give your answer correct to at least two significant figures. This question is worth 1 mark.

Solution This is effectively identical to question A2 from the exercise sheet for Probability 2. Independence of

and

gives that

2. The random variable takes values: 1 with probability 0.6. 4 with probability 0.29. 7 with probability 0.11. Calculate the variance of

.

You should give your answer correct to at least two significant figures. This question is worth 1 mark.

Solution This is effectively identical to question A6 from the exercise sheet for Probability 3.

/

3. The random variable Calculate

. .

You should give your answer correct to at least two significant figures. This question is worth 1 mark.

Solution This is effectively identical to question A2 from the exercise sheet for Probability 4.

4. The random variable Calculate

. .

You should give your answer correct to at least two significant figures. This question is worth 1 mark.

Solution This is effectively identical to question A5 from the exercise sheet for Probability 4.

5. The random variable Calculate

. .

You should give your answer correct to at least two significant figures. This question is worth 1 mark.

/

Solution This is effectively identical to question A3 from the exercise sheet for Probability 5.

6. The random variable

.

Calculate

.

You should give your answer correct to at least two significant figures. This question is worth 1 mark.

Solution This is effectively identical to question A4 from the exercise sheet for Probability 6.

7. The random variable

.

Calculate the value of

such that

.

You should give your answer correct to at least two significant figures. This question is worth 1 mark.

Solution

The final result can be obtained directly using the Stata command “display invnormal(0.9)”.

8. The random variables

are independent and identically distributed, with

Let Calculate

.

denote the average of the 29 random variables. .

You should give your answer correct to at least two significant figures. This question is worth 1 mark.

Solution /

This question is effectively identical to question A1 from the exercise sheet for Probability 6.

9. The number of student emails received by a lecturer in a single day can be considered to be Poisson distributed, with parameter . The lecturer hopes to receive a daily quota of at least one email and at most four emails per day. Less than one email in a day is an indication that the lecturer’s students are not working hard enough, while more than four emails in a day require too much time to respond to. When answering this question you may assume that the number of emails received each day are independent.

a. Calculate the probability that the number of emails received by the lecturer in a single day satisfies their daily quota. You should give you answer correct to at least four significant figures. This part of the question is worth 1 mark.

b. Calculate the probability that across a seven day week, the number of emails received by the lecturer satisfies their daily quota on exactly five of the days. You should give your answer correct to at least two significant figures. This part of the question is worth 2 marks.

Solution Let be the number of emails received in a day. The question therefore states that The daily quota is satisfied if . This occurs with probability

.

Let be the number of days in a seven day week on which the quota is satisfied. Using the result of the first part of the question we then have that . The probability that the quota is satisfied on exactly five days is then

10. The mean daily rainfall in the UK is 2mm. The standard deviation of daily rainfall in the UK is 1.5mm. A 31 day month is declared to be a drought when the total UK rainfall across the 31 days is less than 37mm.

/

When answering this question you may assume that daily rainfall values are independent and identically distributed.

a. Calculate the average UK daily rainfall value below which a 31 day month may be declared to be a drought. You should give your answer correct to at least four significant figures. This part of the question is worth 1 mark.

b. Calculate the probability that a 31 day month is declared to be a drought in the UK. You should give your answer correct to at least two significant figures. This part of the question is worth 2 marks.

Solution Let

be the rainfall on day . A 31 day month is declared to be a drought if

The question then gives that we have that

and

. Across

days the Central Limit Theorem applies, and so

11. A store receives pumpkins from two farms. One third of the pumpkins are supplied by Farm A, with the other two thirds of the pumpkins supplied by Farm B. The weights of pumpkins from Farm A are normally distributed, with mean 4.1kg and standard deviation 0.4kg. The weights of pumpkins from Farm B are normally distributed, with mean 3.9kg and standard deviation 0.3kg. A pumpkin is classified as being underweight if it weighs less than 3.4kg.

a. Calculate the probability that a pumpkin received by the store is underweight, conditional on the knowledge that it was supplied by Farm B. You should give your answer correct to at least four significant figures. This part of the question is worth 1 mark.

b. Calculate the probability that a pumpkin received by the store was supplied by Farm B, conditional on the knowledge that it is underweight. You should give your answer correct to at least two significant figures. This part of the question is worth 2 marks.

Solution /

Let be the weight of the pumpkin, and be the event that the pumpkin is underweight. Let A, and be the event that a pumpkin is from Farm B. If we are focusing on the case where the pumpkin is from Farm B, then

The second part of the question asks for . Bayes’ rule allows us to express in terms of

be the event that a pumpkin is from Farm . We therefore have

which was determined in the first part of the question.

An approach similar to the first part of the question produces and . Combining all of this information together we have

. We also have that

12. An insurance company sells a policy for home insurance. Each month they are required to pay out: £0 with probability 0.7 £250 with probability 0.25 £1000 with probability 0.05 When answering this question you may assume that the monthly payouts are independent.

a. Calculate the mean amount of money paid out in one month. You should give your answer correct to at least four significant figures. This part of the question is worth 1 mark.

b. Calculate the probability that the policy pays out £1000 in total across a period of four months. You should give your answer correct to at least two significant figures. This part of the question is worth 2 marks.

Solution Let be the amount of money paid out in one month. The question then specifies the distribution of as 0

250

1000

0.7

0.25

0.05

This gives that

Let be the total payout over four months. We could attempt to determine the entire distribution of . Instead, we realise that we are only interested in the outcome . A total payout of £1000 is the result of four payments of £250, occurring with probability . Alternatively, £1000 could be paid out in one month and £0 in the remaining three months. There are arrangements of the one £1000 month out of the four possible. The probability of this scenario is therefore Therefore,

....