College Research for Junior Students

a) Let the number of cups sold during the first third of 2017 be represented by letter X. Then,

(956+912+X)/3 = 860

(956+912+X) = 860x3 =2580

X = 712

The number of cups sold during the first third of 2017 is approximately 712

Total marks          /1

f.

The answers in part d and e are unreliable. This is because they are estimates obtained purely by eye and which may be subjective to bias which may be affected by different market forces. The actual figures for the selected periods may be higher or lower. In addition, the estimates above have been calculated without considering the correlation coefficient between sales and thirds. This could have been important in determining the level that the two variables are related.

Total marks          /3

Task 1 total marks:                /16

Task 2

a.

The geometric distribution is given by a probability density function given by:

          f(x) = (1-p)x-1p where x represents the number of attempts and p is the probability of success

Let the probability of passing the test be 0.7 (p=0.7), the probability of failing will be given by 1-p = 0.3

The attempts needed for the probability of first passing to be below 1/500 is;

1/500 › (0.3) x-10.7

1/ (500x0.7) › (0.3) x-1

Log (1/(500x0.7)) › (x-1) log 0.3

x-1 ‹ 4.8655

x ‹ 5.8655 which is approximately 6 attempts

Total marks          /5

b

Let the probability of passing the test be 0.12. The probability of failing the test will be (1-0.12) = 0.88

The number of attempts needed for the probability of first passing to be below 1/50 is:

(0.88) x-10.12 ‹ 1/50

(0.88) x-1 ‹ 1/ (50x0.12)

(x-1) log(0.88) < log (1/ (50x0.12))

x-1 < log (1/ (50x0.12))/(log0.88))

x-1 <14.016

x<15.016, which is approximately 15

Therefore, less than 15 attempts are needed for the probability of the first passing to be below 1/50

Total marks          /5

c.

The answers in part [a] and [b] are based on the assumption that each of the attempts is independent of the other. This means that the outcome of one attempt does not affect the next. The probability of success is the same for each attempt. This outcome is not likely to hold in practice. This is because assuming one fails in the first attempt; it is more likely that they will refer to the mistakes of the first attempt in order to make corrections in a bid to ensure that they pass in the next attempt.

Total marks          /2

Task 2 total marks:                 /12

Task 3

a.

Assuming that there are only 365 birthdays, the probability that two people share the same birthday is 1/365. The probability that no two people share same birthday = 1-1/365 = 364/365

Let k be the number of people one meets. The probability that out of the k people one meets, none of them has a birthday same as his is (364/365)k. Thus, probability of meeting at least one person who has the same birthdate = 1-(364/365)k

Assuming that the probability of meeting this person is more likely than not means that it should be more than 50%

1-(364/365)k

› 50/100

(364/365)k

> 0.5

K log (364/365) > log 0.5

K > 253

Therefore, one needs to meet more than 253 people before it is more likely than not to come across someone whose birthday is on the same date as his.

Total marks         /5

B i. Suppose the four values of n chosen are 10, 20, 40 and 50, the probability of a defect not occurring in a single roll of wallpaper for each value is:

n

Probability of defect

P(n) =(1/n)

Probability of no defect in a single roll (1-P(n))

10

1/10 = 0.1

1-0.1 = 0.9

20

1/20 = 0.05

1-0.05 = 0.95

40

1/40 = 0.025

1-0.025 = 0.975

50

1/50 = 0.02

1-0.02 = 0.98

Total marks          /4

B ii

Number of rolls (n)

Probability of no defect in a single roll (1-p(n))

Probability of a defect not occurring in any of each roll selected (1-P(n))n

10

0.9

0.3487

20

0.95

0.3585

40

0.975

0.3632

50

0.98

0.3642

It was noticed that the probability of a defect not occurring in any of each of n selected rolls are between 0.36 and 0.37

Total marks          /2

B iii

In order to investigate further, more values of n were selected as shown in the table below.

Number of rolls (n)

Probability of defect

P(n) =(1/n)

Probability of no defect in a single roll (1-P(n))

Probability of a defect not occurring in any of each roll selected (1-P(n))n

80

1/80 = 0.0125

0.9875

0.3656

200

1/200 = 0.005

0.995

0.3670

500

1/500 = 0.002

0.998

0.3675

1000

1/1000 = 0.003

0.999

0.3677

20000

1/20000 = 0.00005

0.99995

0.3679

Total marks         /5

Task 3 total marks:              /16

Summary and Conclusion

Task 1 summary:

The process of moving averages in forecasting can be applied in solving different challenges in everyday life including estimation of future sales as in the case of this report

Task2 summary:

There are many probability distributions that can be used in solving different probability problems, each with its assumptions

Task 3 summary:

It was discovered that there are interesting applications and patterns of probability distributions in real life.

Conclusion

To conclude, all the questions of this task have been solved as required.

Summary and conclusion total marks:                 /6