Basics of Probability and Statistics Week 1 Challenge
Q1.
| Sample Variance | Population Variance |
|---|---|
| The formula for sample variance is given as ∑ni=1(xi−μ)2n−1 ∑ i = 1 n ( x i − μ ) 2 n − 1 | The formula for population variance is equal to ∑ni=1(xi−μ)2n ∑ i = 1 n ( x i − μ ) 2 n |
Where xi = ith. Number, x- = mean and n = number of items.
when the variance is to be calculated from population data, n is equal to the number of items. Thus if variance in blood pressure of all the 1000 people is to be calculated from data on blood pressures of all the 1000 people, then n = 1000. However when the variance is calculated from sample data 1 is to be deducted from n before dividing the sum of the squared deviations. Thus in the above example if sample data have 100 items, the denominator would be 100 – 1 = 99.
To put it simply (n−1) is a smaller number than (n). When you divide by a smaller number you get a larger number. Therefore when you divide by (n−1) the sample variance will work out to be a larger number.
Let’s think about what a larger vs. smaller sample variance means. If the sample variance is larger than there is a greater chance that it captures the true population variance. That is why when you divide by (n−1) we call that an unbiased sample estimate. Whereas dividing by (n) is called a biased sample estimate
Population variance refers to the value of variance that is calculated from population data, and sample variance is the variance calculated from sample data. Due to this value of denominator in the formula for variance in case of sample data is ‘n-1’, and it is ‘n’ for population data. As a result both variance and standard deviation derived from sample data are more than those found out from population data
Q2.
we have {M,M,M}{F,F,F}{K,K,K}
cluster sampling M,F,K cluster wise
stratified sampling M,M,K any random
The way in which both systematic and cluster sampling pull sample points from the population is different. While systematic sampling uses fixed intervals from the larger population to create the sample, cluster sampling breaks the population down into different clusters.
Q3.
Difference as sampling from an infinite population (sampling with replacement) versus sampling from a finite population (without replacement).
No. It is N raised to the power n, if you count different orderings as being different samples. If you only count the samples without order, then there are even fewer samples, but it becomes complicated to count, because the number if arrangements of a given sample depends on the pattern of repeats. Thus, each un ordered sample has n! orderings if all are distinct as is the case when drawing without replacement, but for instance the letters aabbbcccc have only 9!/2!3!4! arrangements. In any case, the answer is at most N raised to the power n, which is clearly finite.
Q4.
A deck of cards contains 52 cards. Out of these, there are ]4 Jacks. So,
2/52
2 jacks out of 52 cards
Q5.
dicce have 6 sides so {1,2....6}
and for 2nd dice {1,2,..6}
6*6=36 posiblities total
11 22 33 44 55 66 = 6
hence 6/36=1/6 is the probability
Q6
when we calculate mean the unit is km\hr then ater mean it is same km\hr
in variance we need to sequre it for{ negative and positive dispersion}.
if -2,-1,0,1,2 are dispersion we added it we get zero so we take negative sequre so it became positive .
after sequring units are also sequring km2 \hr2
to remove the sequres S.D. invented `root , veriance`