Basics of Probability and Statistics Week 1 Challenge

1)

Lets us consider sample of 100 items with sample mean of items as 10. So the sum of all the value must be 1000. As result, once we know the value of 99 items, the value of the last item is determined on its own and thus it is not free to vary.The first (n-1) samples are independent of each other given the mean. Thus this will cause a bias in sampled Variance which tends to underestimate the population variance. Thus (n-1) is introduced to to get a unbiased estimate of population variance.

Population Variance =σ² = Σ (xi - μ)² / N where sample variance =s² = Σ (xi - x̄)² / (n-1)

2 )

In stratified sampling the population is seperated into multiple sub-groups called strata based on their homogenous classifications. The sample is selected from strata using random number generator.

Example: A survey that selects candidates based on Gender, Age,country etc.

Clustered sampling is where population is divided into clusters. Here the clusters should be a mini-representaion of the enire pouplation. There should not be any overlaps of the samples.

Example: A Nation wide study which is done on reading capacity of students divded based on schools.

3)

If sampling is done without replacement, meaning that once an individual is selected for a sample, they are removed from the population and cannot be selected again, then the number of different samples of size N that can be selected from a population of size n is given by :

n choose N = n! / (N! * (n-N)!)

If sampling is done with replacement, meaning that an individual can be selected more than once for a sample, then the number of different samples of size N that can be selected from a population of size n is given by the formula:

n^N

4)

The pack contains a total of 52 cards in which we have 4 Jacks.

The probaility of getting one jack is 4/52 =1/13

Now after removing one Jack, we have 51 cards and 3 jacks,

which gives the probability of getting one jack as 3/51=1/17

To find the probability of both events happening (drawing two jacks in a row),

we multiply both events = 1/13*1/17 =1/221

5)

The total number possibilities for 2 dices on any event = 36

The possibility of an event to have same number on both dices is =6

So the probability of two dices having same number is given as 6/36 = 1/6

6)

During the calculation of variance the difference may have negative values which is squared in order to negelect the cancellation of values. This may lead to change in the units of the data which may not be accusate to be used in calculation.In order to avoid the this the Standard Deviation is introduced which Squared Root of variance.

The square root operation used to calculate standard deviation eliminates the effect of extreme outliers on the measure of dispersion, whereas variance is sensitive to outliers

Standard Deviation σ = √Variance