Introduction To Probability And Statistics

Defining Probability Distribution and its Types

1. The probabilities that are possessed by the values of a random variable gives the probability distribution of that particular random variable. A probability distribution can be represented in the form of a function, a table or a graph.

Probabilities possessed by discrete random variables is a discrete probability distribution. In a discrete random variable, the probabilities are assigned to each values and are greater than 0. The sum of all the allotted probabilities is equal to 1. For example, in the tossing of two coins, the probability distribution of obtaining heads is given by the following table:

Number of Heads	0	1	2 Save Time On Research and Writing Hire a Pro to Write You a 100% Plagiarism-Free Paper. Get My Paper	Total
Probability	1/4	2/4 = 1/2	1/4	(1/4) + (1/2) + (1/4) = 1

The probabilities possessed by a continuous random variable is a continuous probability distribution. In a continuous random variable, there are no particular points to which probabilities can be assigned. Thus, the probabilities at each point is zero. However, the area under the range of a curve is denoted by the probability of that range of values. The total area under the probability curve equals to 1.

For example, the waiting time in a bus stop for the arrival of a bus is a continuous random variable and the probability of waiting is the probability distribution of the waiting time.

1. The following table gives a record of the sales over the last 100 days of the top selling bread loaf of a baker.

NUMBER SOLD	NUMBER OF DAYS
0	5
1	15
2	20
3	25
4	20
5	15
Total	100

To calculate the necessary probabilities given the following questions, the following calculations are necessary to be conducted:

NUMBER SOLD	NUMBER OF DAYS	PROBABILITIES
0	5	(5 / 100) = 0.05
1	15	(15 / 100) = 0.15
2	20	(20 / 100) = 0.2
3	25	(25 / 100) = 0.25
4	20	(20 / 100) = 0.2
5	15	(15 / 100) = 0.15
Total	100	(0.05 + 0.15 + 0.2 + 0.25 + 0.2 + 0.15) = 1

1. The required probability that 3 or 4 loaves can be sold in any one day is given by:

P (Selling 3 loaves in one day) * P (Selling 4 loaves in one day) = 0.25 * 0.2 = 0.05.

1. To calculate the average daily sales of the period, the product of the probabilities with the respective number of bread loaves sold has to be calculated. The sum of this product gives the average daily sales. Thus,

The average daily sales = (0 * 0.05) + (1 * 0.15) + (2 * 0.2) + (3 * 0.25) + (4 * 0.2) + (5 * 0.15) = 2.85.

1. The required probability that 2 or more than 2 loaves of bread can be sold on any one day is given by:

P (Selling 2 loaves in one day) + P (Selling 3 loaves in one day) + P (Selling 4 loaves in one day) + P (Selling 5 loaves in one day) = 0.2 + 0.25 + 0.2 + 0.15 = 0.8

1. The required probability that 4 or less loaves of bread can be sold on any one day is given by:

P (Selling 0 loaves in one day) + P (Selling 1 loaf in one day) + P (Selling 2 loaves in one day) + P (Selling 3 loaves in one day) + P (Selling 4 loaves in one day) = 1 – P (Selling 5 loaves in one day) = 1 – 0.15 = 0.85

1. If a coin is tossed, there are two possible outcomes, Head and Tail. Let the event of obtaining a head be denoted by H and the event of obtaining a tail be denoted by T. Now, if a coin is tossed twice, then the number of possible outcomes will be 2²= 4. The four possible outcomes that can be obtained after tossing a coin twice are {HH, HT, TH, TT}. Here, HH denotes both the outcomes are heads, HT denotes that the first outcome is head and the second is tail, TH denotes that the first outcome is a tail and the first is a head and TT denotes both the outcomes are tails.
2. A head on the first toss can be obtained in 2 out of the four cases. Therefore, the required probability of obtaining a head on the first toss is (2 / 4) = 0.5.
3. The number of cases in which tail is obtained on the second toss given that a head has been obtained on the first toss is 1. Thus, the required probability is (1 / 4) = 0.25.
4. The number of cases in which two tails can be obtained is 1 out of 4. Thus, the required probability of obtaining two tails is (1 / 4) = 0.25
5. The number of cases in which tail is obtained on the first toss and a head is obtained on the second toss is 1. Thus, the required probability is (1 / 4) = 0.25.
6. The number of cases in which tail is obtained on the second and a head is obtained on the first toss is 1 out of 4. The number of cases in which tail is obtained on the first toss and a head is obtained on the second toss is 1 out of 4. Therefore, the total number of cases favourable to the event is 2 out of 4. Thus, the required probability is (2 / 4) = 0.5.
7. The number of cases in which at least one head can be obtained within the first two tosses is 3 out of 4. Thus, the required probability is (3 / 4) = 0.75.
8. It is given that the average number of sales of apples is 5000 and the standard deviation of the sales has been obtained as 600. Let the number of apples sold be denoted by X.
9. The probability that the number of apples sold will be greater than 5600 apples is given by P (X > 5600). Now,

 

1. The probability that the number of apples sold will be less than 5240 apples is given by P (X < 5240). Now,

 

1. The probability that the number of apples sold will be less than 4400 apples is given by P (X < 4400). Now,

1. On searching the ABS website, the latest figures on the age and sex of Australian population that has been obtained is given in the following table.

Table 2.1: Male and Female population of Australia in 2017

 

• The probability of selecting a male person at random = (12,204,419 / 24,598,933) = 0.496.
• The probability of selecting a person aged between 55 and 64 years at random = (2,839,984 / 24,598,933) = 0.115.
• Probability of selecting a female between 15 and 24 years of age at random = (1,566,973 / 24,598,933) = 0.064.
• The probability that a person selected at random will be aged 55 years or more = ((2,839,985 + 3,794,800) / 24,598,933)) = 0.270.

• According to the Question,

Mean = 20 hours and Standard deviation = 5 hours.

A random sample of 64 observations were collected.

Confidence level = 95%

• UCL = (Mean + 2 * Standard Error) = 20 + 2 * 1.25 = 22.5

LCL = (Mean – 2 * Standard Error) = 20 – 2 * 1.25 = 17.5

• If the sample size is reduced to 16, then the control limits are given by (Mean A₃ * Standard Deviation), where A₃ is the constant for finding the control limits evaluated from the control limit tables. For a sample of size 16, A₃ is given by 0.739. Therefore,

UCL = (Mean + A₃ * Standard Deviation) = 20 + 0.739 * 5 = 23.7

LCL = (Mean – A₃ * Standard Deviation) = 20 – 0.739 * 5 = 16.3

• H₀: μ 9 against

H₁: μ < 9

Here, μ = 9, n = 50

 = 10.22, σ = 5

df = 50 – 1 = 49

Tabulated value of t for a lower tail test with 49 df = -1.677.

Test Statistic:

 

The value of the test statistic is greater than the tabulated value of t. Thus, H₀ is accepted. Average distance from the store within which the customers stay is more than 9 km.

References

‘3101.0 – Australian Demographic Statistics, Jun 2017’ (Abs.gov.au, 2018) <https://www.abs.gov.au/AUSSTATS/[email protected]/DetailsPage/3101.0Jun%202017?OpenDocument> accessed 14 March 2018

Aidara, Nafy. “Introduction to probability and statistics.” (2018).

Von Mises, R. (2014). Mathematical theory of probability and statistics. Academic Press.

Allen, A. O. (2014). Probability, statistics, and queueing theory. Academic Press.

Bharti, S., & Bharti, B. (2014). Fundamentals of Statistics. Textbook of Hospital Administration, 345.

Gupta, S. C. (2016). Fundamentals of Statistics. Himalaya Publishing House.