Basics of Probability and Statistics Week 1 Challenge
1) Why there is a difference in the formula of variance for population and sample
Sampling process includes the selection of certain number of elements from the whole set of the polulation. Only difference in formula is that the numerator Σ ( x − μ ) ^2 is to be divided by the (number of samples - 1) term in sample variance while in population variance same is to be divided by (number of elements in population). Basic reason behind this is that to compensate the our lack of information being provided.
Sampling process is to be taken in consideration though the values of variance obtained is greater then the actual variance of the population. But will be time and hard cost saving compared to taking in consideration the whole population.
2) Difference between stratified and clustered sampling
| Stratified Sampling | Clustered sampling |
| In stratified sampling, the population is divided into finite number of sub groups/ sub population / strata. And every element of the population should belong to only one strata. Further through the final sampling is completed through "Simple Random Sampling". | As the name suggests, the population is divided in to numerous groups and its sub groups / clusters of same property and characteristics and further through either simple random or systematic random technique sample is finally prepared. |
| Eg: Let us say that there is a population of 30 people, including 20 men and 10 women; and we require 6 men and 5 women in a team of quiz then here we are doing the stratified sampling. That " 6 men and 5 women in a team " basically helps in making simple random sampling. | Eg: Let us say that in India the Electronic Socity executive wants to know the performance of " HP " laptops across country then they will group/ form clusters depending upon cities. WOrk is to be allocated to the supervisor of particular cluster and then through sub-ordinates of supervisor the data can be extracted from clusters. Even states / similar language regions can be parameter to form cluster. |
3) How many different samples can be created out of population of size n
This is basically equivalent to selecting some "r' number of element/s from given set having in total "n" elements in it; selection is without repetition. Which is
nCr = (n!)/[(r!)(n-r)!] .
As the number of elements in sample caries the number of sample thus varies for fixed number of population (n) given.
Suppose a group of 5 friends ( Joe, Sue, Tony, Ivan, Ray) is there, of which we have to prepare a team for quiz containing only 3 members of 5. Then sample space contains [(Joe-Sue-Tony) or (Joe-Sue-Ivan) or (Joe-Sue-Ray) or ( Joe-Tony-Ivan) or (Joe-Tony-Ray) or (Joe-Ivan-Ray) or (Sue-Tony-Ivan) or (Sue-Tony-Ray) or (Sue-Ivan-Ray) or (Tony-Ivan-Ray) === 10 samples.
The above reasult can also be obtained by using nCr = (n!)/[(r!)(n-r)!]
Here, n=5; r=3 : 5C3 = 10
Thus nCr should be answer if each element is taken exactly once while preparing single sample.
4) What is the probability of drawing 2 jacks from a pack of card
Total 52 cards, 13 cards of spade : 13 of leaf : 13 of heart : 13 of diamond :
Each contains exactly one jack card. Total 4 jack cards in a pack of card.
So, Required Probability = `(4C2)/(52C2)` = 0.0045
5) What is the probability that both the numbers on dice are same while rolling 2 dice
Die one = { 1,2,3,4,5,6}
Die two = { 1,2,3,4,5,6}
Favourable outcomes = 6
Total outcomes = 36
Required Probability = 6/36 = 0.167
6) Why standard deviation when variance can measure the dispersion in data
The data which will be given to us will be in a particular units. While calculating variance we are actually doing square of the difference between i th observation and mean of observations. Which ultimately gives us final answer in units to the power two (square of the units) So to determine the exact deviation of data set, we are actually doing the square root of the variance which reduces it to units.
Let us say that we have data set of 5 cars speed : 30 , 40 , 50 , 60 , 70 ( km/hr )
Mean = 50 km/hr
Variance = `Σ ( x − μ ) ^2/ 5 = {( 20 ) ^2 + ( 10 ) ^2 + 0 + ( 10 )^ 2 + ( 20 ) ^2}/ 5` = 200 (km/hr)^2
Standard deviation =√(500) = 14.14 km/hr