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 |
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.
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 |
P (Selling 3 loaves in one day) * P (Selling 4 loaves in one day) = 0.25 * 0.2 = 0.05.
The average daily sales = (0 * 0.05) + (1 * 0.15) + (2 * 0.2) + (3 * 0.25) + (4 * 0.2) + (5 * 0.15) = 2.85.
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
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
Table 2.1: Male and Female population of Australia in 2017
Mean = 20 hours and Standard deviation = 5 hours.
A random sample of 64 observations were collected.
Confidence level = 95%
LCL = (Mean – 2 * Standard Error) = 20 – 2 * 1.25 = 17.5
UCL = (Mean + A3 * Standard Deviation) = 20 + 0.739 * 5 = 23.7
LCL = (Mean – A3 * Standard Deviation) = 20 – 0.739 * 5 = 16.3
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.
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