# Finance, Accounting and Banking Question Answers

**Question A**

Probability distribution refers to the general listing of the results of a study together with any accompanying probabilities which usually lie between 0 and 1 inclusive. The aggregate of the probabilities must equal to 1.

Discrete probability distribution captures variables that are selected from certain items of interest. For example when tossing a coin, one can only obtain one outcome out of the two possible outcomes that is, the head or tail (Nelder, & Baker, 1972).

Continuous probability distributions refer to variables that assume any values that exist at a certain intervals or collections. The intervals are contained in classes that are obtained from grouping of items (Abramowitz, & Stegun, 1965). For example when measuring the height of people, the outcome can take any values due to the nature of the range.

**Question B**

NUMBER SOLD |
NUMBER OF DAYS |
probability |
---|---|---|

0 | 5 | 0.05 |

1 | 15 | 0.15 |

2 | 20 | 0.2 |

3 | 25 | 0.25 |

4 | 20 | 0.2 |

5 | 15 | 0.15 |

Total | 100 | 1.00 |

- The probability of selling 3 or 4 loaves on any one day

P(x=4)

0.25+0.2=0.45

- Average daily sales

0*0.05+1*0.15+2*0.2+3*0.25+4*0.2+5*0.15

=2.85

- P( 2 or more)

1-(0.05+0.15)

=0.8

- P(4 or less)

1-0.15

=0.85

**Question C**

- P(Head)

The probability of obtaining the head is obtained by obtaining the entire possible number of occurrences within a possible study. The coin tossed once has a probability of either the head or the tail appearing. In other words the chances are 50% for the head and 50% for the tail. The probability of the head occurring will be given by the number of possible outcome which is one divided by the total number of occurrences which is 2. Therefore, the probability is as follows;

=0.5

- P(H)*P(T)

The probability of obtaining a head or a tail is subject to conditions. The condition in this case is that the head must be obtained in the first toss. The condition has to be fulfilled for the probability to be obtained. The second toss requires the possibility of obtaining the tail only. Therefore, the probability will be that of obtaining the head in the first toss multiplied by that of obtaining a tail in the second toss.

0.5*0.5

=0.25

- P(T)*P(T)

The probability in this case does not include any conditions. To obtain the probability of obtaining two tails, the coin must be tossed twice. In other words the probability of obtaining the tail in the first toss must be obtained to multiply it with the probability of obtaining the tail in the second toss.

0.5*0.5

=0.25

- P(T)*P(H)

The probability of obtaining the tail in the first toss is obtained by dividing the number of outcome with the possible occurrences. After that the probability of the head is obtained to multiply with to obtain the entire probability.

0.5*0.5

0.25

- P(T)*(H) + P(H)*(T)

The probability is obtained by multiplying the probability of obtaining the head multiplying the probability of obtaining the tail plus the probability of obtaining the head multiplied by the probability of obtaining the tail.

0.5*0.5+0.5*0.5

=0.5

- P(HT)+P(TH)

0.25+0.25

0.5

**Question D**

Here we use the following formula and read the z values from the tables of the normal distribution curve.

- The probability that sales are greater than 5600

Z value=1

The probability is 0.5 – 0.3413=0.1587

- The probability that sales will be less than 5240

Z=0.4

P= 0.1554

The probability is 0.5+0.1554=0.6554

- the probability that sales are less than 4400 apples

Z= -1

P =0.3413

The probability is 0.5-0.3413=0.1587

**Question 2**

**Age group in years and respective female and male population of Australia**

Age group |
Male |
Female |
TOTAL |

0-14 | 2,122,139 | 2,012,670 | 4,134,809 |

15-24 | 1,524,368 | 1,446,663 | 2,971,031 |

25-54 | 4,903,130 | 4,725,976 | 9,629,106 |

55-64 | 1,363,331 | 1,384,036 | 2,747,367 |

65 and above | 1,736,951 | 2,013,149 | 3,750,100 |

TOTAL |
11,649,919 | 11,582,494 | 23,232,413 |

The source is index mundi from the internet

https://www.indexmundi.com/australia/age_structure.html

- The probability that any person selected at random from the population is a male

It will be given by the total number of males in the entire age groups divided by the total population which includes the males and females as follows;

11,649,919/23,234,413

=0.5014

- The probability that any person selected at random from the population is aged between 55 and 64.

The probability will be given by the computation of the total number of males and females in the age group divided by the total population in Australia

2,747,367/23,234,413

=0.1182

- The joint probability that any person selected at random from the population is a female and aged between 15 and 24.

The probability is computed based on the total number of females in the age group divided by the entire population of Australians.

1,446,663/23,234,413

=0.0623

- The probability that any person selected at random from the population is 55 or over.

The probability is computed from the total number of females and males in the age group of 55 and above

(2,747,367+3,750,100)/23,234,413

=0.2796

**Question 3a**

- Appropriate UCL and LCL

=20 ± 1.225

The upper limit is 21.225 while the lower limit is18.775

- If management wishes to use smaller samples of 16 observations calculate the control limits covering the 95% confidence interval.

=20 2.45

The upper limit is 22.45 and the lower limit is 17.55

**Question 3b**

- The null and alternative hypothesis in this case will be as follows;

Ho: µ=9

Ha: µ≤9

- The critical value

It is computed assuming that there is a normal distribution. The population mean is 9 and the sample mean is 10.22. The test statistic to be used in the computation is will be

Therefore, the Z value

The z value calculated is 1.73

The confidence level is 0.05 meaning Z value read from the table at 0.05 is 1.645

9 10.22

In conclusion, the Z calculated is more than the Z critical therefore we reject the null hypothesis. It means that at 95% confidence level the average customer lives in more than 9 KMs.

#### References

Abramowitz, M., & Stegun, I. A. (1965). With formulas, graphs, and mathematical tables. *National Bureau of Standards Applied Mathematics Series e*, *55*, 953.

Nelder, J. A., & Baker, R. J. (1972). *Generalized linear models*. John Wiley & Sons, Inc..