Basics of Probability and Statistics Week 1 Challenge
1.Variance:This term refers to stastical measurement ofspread between the numbers in data set,to be more specified the variance measures how far each number in set from mean and,therefore ,from every other number in set.
Sample Variance:The sample variance is used to calculate how varied the sample is.The sample is the slected number of items from the population i.e from the total number present in a set.The sample variance can be found by using the formula:-
`s2=sigma(x-xbar)/(n-1)`
Where,s2=sample variance
x=value of the element
xbar=sample mean
n=sample size
Population variance:This population variance tells us how the data points in a specific population are spread out.It is an average of distances from each data point in population to the mean,squared.The population variance can be found out by using formula:-
`sigma2=(sigma(x-xbar))/(n)`
Where,sigma2=population variance
x=value of the element
xbar=population mean
n=population size.
As we can from above,both the formula are different because the purpose of this little difference it to get a better and unbiased estimate of the population‘s variance (by dividing by the sample size lowered by one, we compensate for the fact that we are working only with a sample rather than with the whole population).
2.Stratified Sampling:Stratified sampling is a type of sampling method in which the total population is divided into smaller groups or strata to complete the sampling process. The strata is formed based on some common characteristics in the population data. After dividing the population into strata, the researcher randomly selects the sample proportionally.The formula for the stratified sampling is propotionally.
Clustered Sampling:Cluster sampling is a probability sampling technique where researchers divide the population into multiple groups (clusters) for research. Researchers then select random groups with a simple random or systematic random sampling technique for data collection and data analysis.
| Cluster sampling | Stratified sampling |
| Elements of a population are randomly selected to be a part of groups (clusters). | The researcher divides the entire population into even segments (strata). |
| Members from randomly selected clusters are a part of this sample. | Researchers consider individual components of the strata randomly to be a part of sampling units. |
| Researchers maintain homogeneity between clusters. | Researchers maintain homogeneity within the strata. |
| Researchers divide the clusters naturally. | The researchers or statisticians primarily decide the strata division. |
| The key objective is to minimize the cost involved and enhance competence. |
The key objective is to conduct accurate sampling, along with a properly represented population. |
3.If n units are selected in population are, the total number of possible samples are
N/n
So the probability of selecting any one of these samples is
1/(N/n)
Note that a unit can be selected at any one of the n draws. Let i u be the i
th unit selected in the sample.
This unit can be selected in the sample either at first draw, second draw, …, or nth draw.
Let Pi(j) denotes the probability of selection of ui at the jth draw, j = 1,2,...,n. Then
Pj(i)= P1(i)+ P2(i)+ P3(i)+ .....Pn(i)
n times
n/N
Now if u1,u2,....un are the n units selected in the sample, then the probability of their selection is
P(u1,u2,.....un)=P(u1).P(u2),.....,P(un)
Note that when the second unit is to be selected, then there are (n – 1) units left to be selected in the
sample from the population of (N – 1) units. Similarly, when the third unit is to be selected, then there
are (n – 2) units left to be selected in the sample from the population of (N – 2) units and so on.
If P(u1)=n/N,then
P(u2)=(n-1)/(N-1),.....,P(un)=1/(N-n+1)
Thus, 1 n
P(u1,u2,....un)=(n/N).(n-1/N-1).(n-2)/(N-2).....1/(N-n+1)=1/(N/n).
4.If there are 2 Jacks in a pile of 52 cards already, and you add two more Jacks, then your probablilty of drawing out a Jack is 4 out of 54 (or 4/54), and when you divide the numerator and denominator by 2/2 (because it's a factor of 1, so you don't change the equation), you get 2/26.
5.Certainly, it has a 1 in 6 chance of matching the first die (regardless of its value), hence the 1/6 probability.therefore the probability is 1/6.
6.Standard deviation is only used to measure spread or dispersion around the mean of a data set. ... For data with approximately the same mean, the greater the spread, the greater the standard deviation. If all values of a data set are the same, the standard deviation is zero (because each value is equal to the mean).