Unit: Quantitative Analysis
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Login to Access| No use | Light use | Heavy use | ||
| No use | 0.35 | 0.15 | 0.50 | |
| Previous day | Light use | 0.30 | 0.35 | 0.35 |
| Heavy use | 0.15 | 0.30 | 0.55 |
| Sales Sh.“000” | Probability of outstanding invoice |
| 0 – 100 | 0.20 |
| 100 – 200 | 0.18 |
| 200 – 300 | 0.22 |
| 300 – 400 | 0.16 |
| 400 – 500 | 0.09 |
| 500 – 600 | 0.08 |
| 600 – 700 | 0.04 |
| 700 – 800 | 0.03 |
Required:
Using normal distribution approach:
(i) Calculate the value that will be exceeded 10% of the time assuming an observation is randomly selected from the distribution.
(ii) Compute the value that will be exceeded 85% of the time assuming an observation is randomly selected from the distribution.
(iii) Determine the two values of which the smaller value has 25% of the values below it and the larger value has 25% of the values above it.
(iv) Determine the value in which 15% of the observations will be below the distribution.
(i) Transition matrix.
(ii) Equilibrium state.
(iii) Initial probability vector.
Required:
The probability that the output of an acre of land:
(i) Is greater than 48 bags.
(ii) Is greater than 60 bags.
(iii) Is less than 45 bags.
(iv) Lies between 50 bags and 60 bags.
| Result | % of candidates |
| Passed with distinction | 10% |
| Passed with credit | 60% |
| Failed | 30% |
| Opinion | |||
| Students gender | In favour | Opposed | Undecided |
| Male | 40 | 10 | 15 |
| Female | 20 | 30 | 20 |
| Profit Sh.“million” | Probability |
| 10 – 20 | 0.05 |
| 20 – 30 | 0.05 |
| 30 – 40 | 0.10 |
| 40 – 50 | 0.15 |
| 50 – 60 | 0.30 |
| 60 – 70 | 0.10 |
| 70 – 80 | 0.20 |
| 80 – 90 | 0.05 |
(i) Joint probability.
(ii) Mutually exclusive events.
(iii) Conditional probability.
(iv) Dependent events.
Required:
(a) The probability that one customer will have drawn five defective components by the end of 5 weeks.
(b) The probability that two customers will have drawn two defective components each, two none and the other components, in two weeks.
Required:
Using the Poisson distribution, calculate the approximate number of packets containing:
(i) No defectives.
(ii) 1 defective.
(iii) 2 defectives.
(i) Binomial distribution.
(ii) Poisson distribution.
| Type of defect | Probability |
| A | 0.15 |
| B | 0.14 |
| C(if it neither contains defect A nor defect B) | 0.3 |
| C(if it contains either defect A or defect B) | 0.2 |
| C(if it contains both defects A and B) | 0.1 |
|
Digital boxes sold (units) Probability |
0 0.05 |
1 0.05 |
2 0.10 |
3 0.15 |
4 0.20 |
5 0.15 |
6 0.15 |
7 0.10 |
8 0.05 |