Basics of Probability and Statistics Week 1 Challenge
1)
Variance formla for population= `\Sigma \frac{(x - \mu )^2}{\N}`
Variance formula for sample = `\Sigma \frac{(x - \mu )^2}{\N-1}`
The difference in formulas of Variance for population and variance of sample is (N-1). If we take some sample data to calculate the sample variance, it will give accurate answer than the population variance as population comprises of overall data whereas sample is a part of population.
2)
In stratified sampling gross are chosen specifically to represent different characteristics within the population, such as ethnicity, location,age or occupation. Within each group a random sample is taken, sometimes in proportion to the size of the group.
Stratified sampling can lead to a very good random representative sampling, unbiased.
In cluster sampling the population is divided into clusters which are then chosen at random. For example department of a business can be clusters . Within each cluster all of the objects are included in the sample. Cluster sampling can be more convenient and practical.
However if clusters are different from each other with regards to the element we are measuring it can lead to bias or non-representativeness
3)
Here the 'n' is the size of the population and it is not the defined number to take the samples. So if we consider "N" as the size of certain subgroup data then "N1" will be another subgroup data size and so on.
Hence we can formulate as many as N1,N2,N3 as many data samples as we want.
4)
Probability of drawing 2 jacks from a pack of card
Prob of getting 2 cards from 52 :-
52C2 = 1326
Prob of getting 2 jacks from 4 :-
4C2 = 6
So our probability of getting 2 jacks from a pack of card is = 6/1326
5)
As there are 2 dices the no of possibilities we get is (6)^2 = 36
No of equal numbers possibilities are (1,1),(2,2),(3,3),(4,4),(5,5),(6,6)
Hence the no of occurances are 6
The probability that both the numbers on dice are same while rolling 2 dice is = 6/36 => 1/6
6)
To calculate the dispersion with more accuracy, standard deviation gives better results than variance. Standard deviation observes how spread out a group of numbers is from the mean, while the variance measures the average degree to which each point differs from the mean.
Let us consider a small data of marks obtined by 3 students out of 100 = [90,85,100]
We know standard deviation-
`\sigma = \sqrt{\frac{ \Sigma (x - \mu )^2}{N} }`
`\mu = \Sigma (\frac{x}{N})``
`mu = (90+85+100)/3`
\mu = 91.6
\sigma_{1} = \sqrt{ \frac{(90-91.6)^2}{3} } = 0.94
\sigma_{2} = \sqrt{ \frac{(85-91.6)^2}{3} } = 3.88
\sigma_{3} = \sqrt{ \frac{(100-91.6)^2}{3} } = 4.94
Also variance -
S^2 = \frac{(x - \mu )^2}{N-1}
S_{1} ^2 = \frac{(90 - 91.6 )^2}{3-1} = 1.28
S_{2} ^2 = \frac{(85 - 91.6 )^2}{3-1} = 21.78
S_{3} ^2 = \frac{(100 - 91.6 )^2}{3-1} = 37.84
From the aove results we can find that to calculate dispersion, we get better accuracy at standard deviation than at variance.