Free BECC-107 Solved Assignment | July 2023-January 2024 |STATISTICAL METHODS FOR ECONOMICS | IGNOU

BECC-107 Solved Assignment

For July 2023 and January 2024 Admission Cycles

Answer the following Descriptive Category Questions in about 500 words each. Each question carries 2 0 2 0 20\mathbf{2 0}20 marks. Word limit does not apply in the case of numerical questions.
  1. (a) Calculate mean, median and mode from the following data.
Class Interval Frequency
3 4 3 4 3-43-434 3
4 5 4 5 4-54-545 7
5 6 5 6 5-65-656 22
6 7 6 7 6-76-767 60
7 8 7 8 7-87-878 85
8 9 8 9 8-98-989 32
9 10 9 10 9-109-10910 8
Class Interval Frequency 3-4 3 4-5 7 5-6 22 6-7 60 7-8 85 8-9 32 9-10 8| Class Interval | Frequency | | :—: | :—: | | $3-4$ | 3 | | $4-5$ | 7 | | $5-6$ | 22 | | $6-7$ | 60 | | $7-8$ | 85 | | $8-9$ | 32 | | $9-10$ | 8 |
(b) Calculate the coefficient of variation from the data given above.
2. Bring out the distinction between sample survey and census. Describe the steps you would follow in collecting data though a sample survey. Prepare a small questionnaire for collection of income and expenditure levels of households.

Assignment II

Answer the following Middle Category Questions in about 250 words each. Each question carries 1 0 1 0 10\mathbf{1 0}10 marks. Word limit does not apply in the case of numerical questions.
  1. a) The probability that Rajesh will score more than 90 marks in class test is 0.75 . What is the probability that Rajesh will secure more than 90 marks in three out of four class tests?
    b) Bring out the major properties of binomial distribution. Mention certain important uses of this distribution.
  2. a) Fit a straight line ( Y = a + b X ) ( Y = a + b X ) (Y=a+bX)(Y=a+b X)(Y=a+bX) to the following data. Compare the estimated values of the dependent variable with its actual values.
X 5 8 10 12 13 15 17 16
Y 8 12 14 10 13 16 14 17
X 5 8 10 12 13 15 17 16 Y 8 12 14 10 13 16 14 17| X | 5 | 8 | 10 | 12 | 13 | 15 | 17 | 16 | | :— | :— | :— | :— | :— | :— | :— | :— | :— | | Y | 8 | 12 | 14 | 10 | 13 | 16 | 14 | 17 |
b) Define correlation coefficient. What are its properties?
  1. What is a life table? Explain its uses and limitations.

Assignment III

Answer the following Short Category Questions. Each question carries 1 5 1 5 15\mathbf{1 5}15 marks.
  1. Write short notes on the following:
    (a) Bayes’ theorem of probability
    (b) Age specific birth and death rates
    (c) Measurement of Skewness
  2. Differentiate between the following:
    (a) Simple random sampling and Stratified random sampling
    (b) Type I and Type II errors in hypothesis testing
    (c) Estimator and Estimate

Expert Answer

BECC-107 Solved Assignment

For July 2023 and January 2024 Admission Cycles

Question:-01(a)

Calculate mean, median and mode from the following data.
Class Interval Frequency
3-4 3
4-5 7
5-6 22
6-7 60
7-8 85
8-9 32
9-10 8
Class Interval Frequency 3-4 3 4-5 7 5-6 22 6-7 60 7-8 85 8-9 32 9-10 8| Class Interval | Frequency | | :—: | :—: | | 3-4 | 3 | | 4-5 | 7 | | 5-6 | 22 | | 6-7 | 60 | | 7-8 | 85 | | 8-9 | 32 | | 9-10 | 8 |

Answer:

Class ( 1 ) Frequency ( f ) ( 2 ) Mid value ( x ) ( 3 ) f x ( 4 ) = ( 2 ) × ( 3 ) c f ( 6 ) 3 4 3 3.5 10.5 3 4 5 7 4.5 31.5 10 5 6 22 5.5 121 32 6 7 60 6.5 390 92 7 8 85 7.5 637.5 177 8 9 32 8.5 272 209 9 10 8 9.5 76 217 n = 2 1 7 f x = 1 5 3 8 . 5 Mean x ¯ = f x n = 1538.5 217 = 7.0899  Class  ( 1 )  Frequency  ( f ) ( 2 )  Mid value  ( x ) ( 3 ) f x ( 4 ) = ( 2 ) × ( 3 ) c f ( 6 ) 3 4 3 3.5 10.5 3 4 5 7 4.5 31.5 10 5 6 22 5.5 121 32 6 7 60 6.5 390 92 7 8 85 7.5 637.5 177 8 9 32 8.5 272 209 9 10 8 9.5 76 217 n = 2 1 7 f x = 1 5 3 8 . 5  Mean  x ¯ = f x n = 1538.5 217 = 7.0899 {:[{:[{:[” Class “],[(1)]:},{:[” Frequency “(f)],[(2)]:},{:[” Mid value “(x)],[(3)]:},{:[f*x],[(4)=(2)xx(3)]:},{:[cf],[(6)]:}],[3-4,3,3.5,10.5,3],[4-5,7,4.5,31.5,10],[5-6,22,5.5,121,32],[6-7,60,6.5,390,92],[7-8,85,7.5,637.5,177],[8-9,32,8.5,272,209],[9-10,8,9.5,76,217],[–,–,—,–,—],[–,n=217,–,sum f*x=1538.5,–]:}],[{:[” Mean ” bar(x)=(sum fx)/(n)],[=(1538.5)/(217)],[=7.0899]:}]:}\begin{aligned} &\begin{array}{|c|c|c|c|c|} \hline \begin{array}{c} \text { Class } \\ \mathbf{( 1 )} \end{array} & \begin{array}{c} \text { Frequency }(\boldsymbol{f}) \\ (\mathbf{2}) \end{array} & \begin{array}{c} \text { Mid value }(\boldsymbol{x}) \\ (\mathbf{3}) \end{array} & \begin{array}{c} \boldsymbol{f} \cdot \boldsymbol{x} \\ (\mathbf{4})=(\mathbf{2}) \times(\mathbf{3}) \end{array} & \begin{array}{c} \boldsymbol{c} \boldsymbol{f} \\ \mathbf{( 6 )} \end{array} \\ \hline 3-4 & 3 & 3.5 & 10.5 & 3 \\ \hline 4-5 & 7 & 4.5 & 31.5 & 10 \\ \hline 5-6 & 22 & 5.5 & 121 & 32 \\ \hline 6-7 & 60 & 6.5 & 390 & 92 \\ \hline 7-8 & 85 & 7.5 & 637.5 & 177 \\ \hline 8-9 & 32 & 8.5 & 272 & 209 \\ \hline 9-10 & 8 & 9.5 & 76 & 217 \\ \hline– & — & — & — & — \\ \hline– & \boldsymbol{n = 2 1 7} & — & \sum \boldsymbol{f} \cdot \boldsymbol{x}=\mathbf{1 5 3 8 . 5} & — \\ \hline \end{array}\\ &\begin{aligned} & \text { Mean } \bar{x}=\frac{\sum f x}{n} \\ & =\frac{1538.5}{217} \\ & =7.0899 \end{aligned} \end{aligned} Class (1) Frequency (f)(2) Mid value (x)(3)fx(4)=(2)×(3)cf(6)3433.510.534574.531.51056225.51213267606.53909278857.5637.517789328.527220991089.576217n=217fx=1538.5 Mean x¯=fxn=1538.5217=7.0899
To find Median Class
= = === value of ( n 2 ) th n 2 th  ((n)/(2))^(“th “)\left(\frac{n}{2}\right)^{\text {th }}(n2)th  observation
= = === value of ( 217 2 ) th 217 2 th  ((217)/(2))^(“th “)\left(\frac{217}{2}\right)^{\text {th }}(2172)th  observation
= = === value of 108 th 108 th  108^(“th “)108^{\text {th }}108th  observation
From the column of cumulative frequency c f c f cfc fcf, we find that the 108 th 108 th  108^(“th “)108^{\text {th }}108th  observation lies in the class 7 8 7 8 7-87-878.
:.\therefore The median class is 7 8 7 8 7-87-878.
Now,
L = L = :.L=\therefore L=L= lower boundary point of median class = 7 = 7 =7=7=7
n = n = :.n=\therefore n=n= Total frequency = 217 = 217 =217=217=217
c f = c f = :.cf=\therefore c f=cf= Cumulative frequency of the class preceding the median class = 92 = 92 =92=92=92
f = f = :.f=\therefore f=f= Frequency of the median class = 85 = 85 =85=85=85
c = c = :.c=\therefore c=c= class length of median class = 1 = 1 =1=1=1
Median M = L + n 2 c f f c M = L + n 2 c f f c M=L+((n)/(2)-cf)/(f)*cM=L+\frac{\frac{n}{2}-c f}{f} \cdot cM=L+n2cffc
= 7 + 108.5 92 85 1 = 7 + 16.5 85 1 = 7 + 0.1941 = 7.1941 = 7 + 108.5 92 85 1 = 7 + 16.5 85 1 = 7 + 0.1941 = 7.1941 {:[=7+(108.5-92)/(85)*1],[=7+(16.5)/(85)*1],[=7+0.1941],[=7.1941]:}\begin{aligned} & =7+\frac{108.5-92}{85} \cdot 1 \\ & =7+\frac{16.5}{85} \cdot 1 \\ & =7+0.1941 \\ & =7.1941 \end{aligned}=7+108.592851=7+16.5851=7+0.1941=7.1941
To find Mode Class
Here, maximum frequency is 85 .
:.\therefore The mode class is 7 8 7 8 7-87-878.
L = L = :.L=\therefore L=L= lower boundary point of mode class = 7 = 7 =7=7=7
f 1 = f 1 = :.f_(1)=\therefore f_1=f1= frequency of the mode class = 85 = 85 =85=85=85
f 0 = f 0 = :.f_(0)=\therefore f_0=f0= frequency of the preceding class = 60 = 60 =60=60=60
f 2 = f 2 = :.f_(2)=\therefore f_2=f2= frequency of the succedding class = 32 = 32 =32=32=32
c = c = :.c=\therefore c=c= class length of mode class = 1 = 1 =1=1=1
Z = L + ( f 1 f 0 2 f 1 f 0 f 2 ) c Z = L + f 1 f 0 2 f 1 f 0 f 2 c Z=L+((f_(1)-f_(0))/(2*f_(1)-f_(0)-f_(2)))*cZ=L+\left(\frac{f_1-f_0}{2 \cdot f_1-f_0-f_2}\right) \cdot cZ=L+(f1f02f1f0f2)c
= 7 + ( 85 60 2 85 60 32 ) 1 = 7 + 85 60 2 85 60 32 1 =7+((85-60)/(2*85-60-32))*1=7+\left(\frac{85-60}{2 \cdot 85-60-32}\right) \cdot 1=7+(85602856032)1
= 7 + ( 25 78 ) 1 = 7 + 25 78 1 =7+((25)/(78))*1=7+\left(\frac{25}{78}\right) \cdot 1=7+(2578)1
= 7 + 0.3205 = 7 + 0.3205 =7+0.3205=7+0.3205=7+0.3205
= 7.3205 = 7.3205 =7.3205=7.3205=7.3205




Question:-01(b)

Calculate the coefficient of variation from the data given above.

Answer:

Class (1) Frequency ( f ) (2) Mid value ( x ) ( 3 ) f x (4) = ( 2 ) × ( 3 ) f x 2 = ( f x ) × ( x ) ( 5 ) = ( 4 ) × ( 3 ) 3 4 3 3.5 10.5 36.75 4 5 7 4.5 31.5 141.75 5 6 22 5.5 121 665.5 6 7 60 6.5 390 2535 7 8 85 7.5 637.5 4781.25 8 9 32 8.5 272 2312 9 10 8 9.5 76 722 n = 2 1 7 f x = 1 5 3 8 . 5 f x 2 = 1 1 1 9 4 . 2 5 Mean x ¯ = f x n = 1538.5 217 = 7.0899  Class   (1)   Frequency  ( f )  (2)   Mid value  ( x ) ( 3 ) f x  (4)  = ( 2 ) × ( 3 ) f x 2 = ( f x ) × ( x ) ( 5 ) = ( 4 ) × ( 3 ) 3 4 3 3.5 10.5 36.75 4 5 7 4.5 31.5 141.75 5 6 22 5.5 121 665.5 6 7 60 6.5 390 2535 7 8 85 7.5 637.5 4781.25 8 9 32 8.5 272 2312 9 10 8 9.5 76 722 n = 2 1 7 f x = 1 5 3 8 . 5 f x 2 = 1 1 1 9 4 . 2 5  Mean  x ¯ = f x n = 1538.5 217 = 7.0899 {:[{:[{:[” Class “],[” (1) “]:},{:[” Frequency “(f)],[” (2) “]:},{:[” Mid value “(x)],[(3)]:},{:[f*x],[” (4) “=(2)xx(3)]:},{:[f*x^(2)=(f*x)xx(x)],[(5)=(4)xx(3)]:}],[3-4,3,3.5,10.5,36.75],[4-5,7,4.5,31.5,141.75],[5-6,22,5.5,121,665.5],[6-7,60,6.5,390,2535],[7-8,85,7.5,637.5,4781.25],[8-9,32,8.5,272,2312],[9-10,8,9.5,76,722],[–,–,—,–,–],[–,n=217,–,sum f*x=1538.5,sum f*x^(2)=11194.25]:}],[” Mean ” bar(x)=(sum fx)/(n)],[{:[=(1538.5)/(217)],[=7.0899]:}]:}\begin{aligned} &\begin{array}{|c|c|c|c|c|} \hline \begin{array}{c} \text { Class } \\ \text { (1) } \end{array} & \begin{array}{c} \text { Frequency }(\boldsymbol{f}) \\ \text { (2) } \end{array} & \begin{array}{c} \text { Mid value }(\boldsymbol{x}) \\ \mathbf{( 3 )} \end{array} & \begin{array}{c} \boldsymbol{f} \cdot \boldsymbol{x} \\ \text { (4) }=\mathbf{( 2 )} \times(\mathbf{3}) \end{array} & \begin{array}{c} \boldsymbol{f} \cdot \boldsymbol{x}^{\mathbf{2}}=(\boldsymbol{f} \cdot \boldsymbol{x}) \times(\boldsymbol{x}) \\ \mathbf{( 5 )}=\mathbf{( 4 )} \times(\mathbf{3}) \end{array} \\ \hline 3-4 & 3 & 3.5 & 10.5 & 36.75 \\ \hline 4-5 & 7 & 4.5 & 31.5 & 141.75 \\ \hline 5-6 & 22 & 5.5 & 121 & 665.5 \\ \hline 6-7 & 60 & 6.5 & 390 & 2535 \\ \hline 7-8 & 85 & 7.5 & 637.5 & 4781.25 \\ \hline 8-9 & 32 & 8.5 & 272 & 2312 \\ \hline 9-10 & 8 & 9.5 & 76 & 722 \\ \hline– & — & — & — & — \\ \hline– & \boldsymbol{n}=\mathbf{2 1 7} & — & \sum \boldsymbol{f} \cdot \boldsymbol{x}=\mathbf{1 5 3 8 . 5} & \sum \boldsymbol{f} \cdot \boldsymbol{x}^{\mathbf{2}}=\mathbf{1 1 1 9 4 . 2 5} \\ \hline \end{array}\\ &\text { Mean } \bar{x}=\frac{\sum f x}{n}\\ &\begin{aligned} & =\frac{1538.5}{217} \\ & =7.0899 \end{aligned} \end{aligned} Class  (1)  Frequency (f) (2)  Mid value (x)(3)fx (4) =(2)×(3)fx2=(fx)×(x)(5)=(4)×(3)3433.510.536.754574.531.5141.7556225.5121665.567606.5390253578857.5637.54781.2589328.5272231291089.576722n=217fx=1538.5fx2=11194.25 Mean x¯=fxn=1538.5217=7.0899
Population Standard deviation σ = f x 2 ( f x ) 2 n n σ = f x 2 f x 2 n n sigma=sqrt((sum f*x^(2)-((sum f*x)^(2))/(n))/(n))\sigma=\sqrt{\frac{\sum f \cdot x^2-\frac{\left(\sum f \cdot x\right)^2}{n}}{n}}σ=fx2(fx)2nn
= 11194.25 ( 1538.5 ) 2 217 217 = 11194.25 10907.7523 217 = 286.4977 217 = 1.3203 = 1.149 = 11194.25 ( 1538.5 ) 2 217 217 = 11194.25 10907.7523 217 = 286.4977 217 = 1.3203 = 1.149 {:[=sqrt((11194.25-((1538.5)^(2))/(217))/(217))],[=sqrt((11194.25-10907.7523)/(217))],[=sqrt((286.4977)/(217))],[=sqrt1.3203],[=1.149]:}\begin{aligned} & =\sqrt{\frac{11194.25-\frac{(1538.5)^2}{217}}{217}} \\ & =\sqrt{\frac{11194.25-10907.7523}{217}} \\ & =\sqrt{\frac{286.4977}{217}} \\ & =\sqrt{1.3203} \\ & =1.149 \end{aligned}=11194.25(1538.5)2217217=11194.2510907.7523217=286.4977217=1.3203=1.149
Coefficient of Variation (Population) = σ x ¯ 100 % = 1.149 7.0899 100 % = 16.21 %  Coefficient of Variation (Population)  = σ x ¯ 100 % = 1.149 7.0899 100 % = 16.21 % {:[” Coefficient of Variation (Population) “=(sigma)/(( bar(x)))*100%],[=(1.149)/(7.0899)*100%],[=16.21%]:}\begin{aligned} & \text { Coefficient of Variation (Population) }=\frac{\sigma}{\bar{x}} \cdot 100 \% \\ & =\frac{1.149}{7.0899} \cdot 100 \% \\ & =16.21 \% \end{aligned} Coefficient of Variation (Population) =σx¯100%=1.1497.0899100%=16.21%




Question:-02

Bring out the distinction between sample survey and census. Describe the steps you would follow in collecting data though a sample survey. Prepare a small questionnaire for collection of income and expenditure levels of households.

Answer:

Distinction Between Sample Survey and Census

Census and sample surveys are two key methods used for collecting data from a population. The primary distinction lies in the scope of the data collection.

1. Census:

  • Definition: A census is a method of data collection in which every unit or member of a population is surveyed. It aims to cover the entire population, leaving no one out.
  • Characteristics:
    • Complete enumeration: Every individual or unit in the population is included.
    • Accuracy: Since it covers the entire population, the data obtained can be highly accurate, assuming no measurement errors occur.
    • Costly and Time-Consuming: Due to its large scale, conducting a census requires significant time, financial resources, and personnel.
    • Infrequent: Censuses are typically conducted at long intervals (e.g., every 10 years) due to their complexity.
  • Example: National population census where every household in the country is surveyed to obtain demographic data.

2. Sample Survey:

  • Definition: A sample survey collects data from a subset of the population rather than the entire population. The subset, or sample, is chosen to be representative of the population.
  • Characteristics:
    • Partial enumeration: Only a sample, not the entire population, is surveyed.
    • Cost-effective: Sample surveys are generally less expensive and quicker to conduct than censuses.
    • Potential for Sampling Error: Since not everyone is surveyed, the results may have some degree of sampling error, though this can be minimized with good sampling techniques.
    • Frequent: Sample surveys can be conducted more frequently due to their lower cost and faster execution.
  • Example: A survey of 1,000 households to determine average household income across a country.

Steps in Collecting Data Through a Sample Survey

To conduct a sample survey efficiently, the following steps should be taken:

1. Define the Objectives of the Survey:

  • Clearly identify the purpose of the survey and the kind of information you want to collect. For instance, you might aim to assess the income and expenditure patterns of households in a particular region.

2. Define the Population:

  • Determine the target population for your survey. This could be the entire population of a city, a specific age group, or households with certain characteristics.

3. Select a Sampling Method:

  • Choose an appropriate sampling method to ensure that the sample is representative of the population. Common methods include:
    • Simple random sampling: Every individual in the population has an equal chance of being selected.
    • Stratified sampling: The population is divided into strata (e.g., income levels), and a random sample is taken from each stratum.
    • Cluster sampling: The population is divided into clusters (e.g., neighborhoods), and entire clusters are randomly selected.

4. Determine the Sample Size:

  • Decide how many individuals or units to survey. The sample size should be large enough to be representative but small enough to be manageable within available resources.

5. Design the Questionnaire:

  • Create a well-structured questionnaire that collects the necessary data. The questions should be clear, concise, and relevant to the objectives of the survey.

6. Conduct a Pilot Survey:

  • Test the questionnaire and survey process on a small sample to identify any potential issues or misunderstandings.

7. Collect the Data:

  • Administer the survey to the selected sample. This can be done through face-to-face interviews, telephone interviews, mailed questionnaires, or online surveys.

8. Process and Analyze the Data:

  • After collecting the data, process it (e.g., coding responses, checking for errors) and analyze it using statistical methods to draw conclusions about the population.

9. Report the Findings:

  • Present the findings in a clear and understandable manner, typically in the form of a report or presentation.

Sample Questionnaire for Collecting Income and Expenditure Levels of Households

Title: Household Income and Expenditure Survey
Instructions: Please answer the following questions honestly and to the best of your knowledge. All information provided will be kept confidential and used only for research purposes.

Section A: Household Information
  1. Number of Household Members: _______
  2. Location of Household (City, Town, Village): ____________________
  3. Main Source of Household Income (Tick one):
    • [ ] Salaried Employment
    • [ ] Self-Employment
    • [ ] Agriculture
    • [ ] Pension
    • [ ] Other: ____________________

Section B: Household Income
  1. What is your household’s total monthly income?
    • [ ] Less than $500
    • [ ] $500 – $999
    • [ ] $1,000 – $1,499
    • [ ] $1,500 – $1,999
    • [ ] $2,000 and above
  2. Does your household receive any additional income from the following sources? (Tick all that apply)
    • [ ] Rental income
    • [ ] Government assistance
    • [ ] Interest or dividends
    • [ ] Remittances from family abroad
    • [ ] Other: ____________________

Section C: Household Expenditure
  1. What is your household’s total monthly expenditure?
    • [ ] Less than $500
    • [ ] $500 – $999
    • [ ] $1,000 – $1,499
    • [ ] $1,500 – $1,999
    • [ ] $2,000 and above
  2. Please estimate your household’s monthly expenditure on the following categories:
Category Monthly Expenditure ($)
Food and groceries ___________________________
Housing (rent/mortgage) ___________________________
Utilities (electricity, water, gas) ___________________________
Education ___________________________
Healthcare ___________________________
Transportation ___________________________
Entertainment and leisure ___________________________
Other ___________________________

Section D: Savings and Investments
  1. Does your household save any portion of its income?
    • [ ] Yes
    • [ ] No
  2. If yes, what percentage of your household’s income is saved monthly?
    • [ ] Less than 5%
    • [ ] 5% – 10%
    • [ ] 11% – 15%
    • [ ] More than 15%
  3. Does your household invest in any of the following? (Tick all that apply)
    • [ ] Real estate
    • [ ] Stocks or bonds
    • [ ] Mutual funds
    • [ ] Pension funds
    • [ ] Other: ____________________

End of Questionnaire
Thank you for your participation!

This questionnaire is designed to gather basic information on household income and expenditure levels, which can be used for analyzing consumption patterns, savings rates, and investment behavior across households.




Assignment II

Question:-03(a)

The probability that Rajesh will score more than 90 marks in a class test is 0.75. What is the probability that Rajesh will secure more than 90 marks in three out of four class tests?

Answer:

To solve this problem, we can model it using a binomial distribution because we are dealing with a series of independent events (class tests), each of which has two possible outcomes: Rajesh scores more than 90 marks (success) or Rajesh scores 90 or fewer marks (failure).

Given Data:

  • The probability that Rajesh scores more than 90 marks in a single test, p p ppp, is 0.75 0.75 0.750.750.75.
  • The number of class tests, n n nnn, is 4.
  • We are asked to find the probability that Rajesh scores more than 90 marks in exactly 3 out of 4 tests, which means k = 3 k = 3 k=3k = 3k=3.

Binomial Probability Formula:

The probability of getting exactly k k kkk successes in n n nnn independent trials is given by the binomial probability formula:
P ( X = k ) = ( n k ) p k ( 1 p ) n k P ( X = k ) = ( n k ) p k ( 1 p ) n k P(X=k)=((n)/(k))*p^(k)*(1-p)^(n-k)P(X = k) = \binom{n}{k} \cdot p^k \cdot (1 – p)^{n – k}P(X=k)=(nk)pk(1p)nk
Where:
  • ( n k ) ( n k ) ((n)/(k))\binom{n}{k}(nk) is the binomial coefficient, representing the number of ways to choose k k kkk successes from n n nnn trials.
  • p p ppp is the probability of success in a single trial.
  • 1 p 1 p 1-p1 – p1p is the probability of failure in a single trial.
  • X X XXX is the number of successes.

Step-by-Step Calculation:

  1. n = 4 n = 4 n=4n = 4n=4, k = 3 k = 3 k=3k = 3k=3, and p = 0.75 p = 0.75 p=0.75p = 0.75p=0.75.
  2. The binomial coefficient ( 4 3 ) ( 4 3 ) ((4)/(3))\binom{4}{3}(43) is calculated as:
( 4 3 ) = 4 ! 3 ! ( 4 3 ) ! = 4 ( 4 3 ) = 4 ! 3 ! ( 4 3 ) ! = 4 ((4)/(3))=(4!)/(3!(4-3)!)=4\binom{4}{3} = \frac{4!}{3!(4 – 3)!} = 4(43)=4!3!(43)!=4
  1. Substituting the values into the binomial probability formula:
P ( X = 3 ) = ( 4 3 ) ( 0.75 ) 3 ( 1 0.75 ) 4 3 P ( X = 3 ) = ( 4 3 ) ( 0.75 ) 3 ( 1 0.75 ) 4 3 P(X=3)=((4)/(3))*(0.75)^(3)*(1-0.75)^(4-3)P(X = 3) = \binom{4}{3} \cdot (0.75)^3 \cdot (1 – 0.75)^{4 – 3}P(X=3)=(43)(0.75)3(10.75)43
P ( X = 3 ) = 4 ( 0.75 ) 3 ( 0.25 ) 1 P ( X = 3 ) = 4 ( 0.75 ) 3 ( 0.25 ) 1 P(X=3)=4*(0.75)^(3)*(0.25)^(1)P(X = 3) = 4 \cdot (0.75)^3 \cdot (0.25)^1P(X=3)=4(0.75)3(0.25)1
  1. Calculate the powers of 0.75 and 0.25:
P ( X = 3 ) = 4 0.421875 0.25 P ( X = 3 ) = 4 0.421875 0.25 P(X=3)=4*0.421875*0.25P(X = 3) = 4 \cdot 0.421875 \cdot 0.25P(X=3)=40.4218750.25
  1. Perform the multiplication:
P ( X = 3 ) = 4 0.10546875 = 0.421875 P ( X = 3 ) = 4 0.10546875 = 0.421875 P(X=3)=4*0.10546875=0.421875P(X = 3) = 4 \cdot 0.10546875 = 0.421875P(X=3)=40.10546875=0.421875

Conclusion:

The probability that Rajesh will secure more than 90 marks in exactly 3 out of 4 class tests is 0.4219 (rounded to four decimal places).




Question:-03(b)

Bring out the major properties of binomial distribution. Mention certain important uses of this distribution.

Answer:

Major Properties of Binomial Distribution

The binomial distribution is one of the most commonly used probability distributions in statistics. It describes the number of successes in a fixed number of independent trials, where each trial has two possible outcomes: success or failure.
Here are the major properties of the binomial distribution:
  1. Fixed Number of Trials (n):
    • The binomial distribution is based on a fixed number of trials, denoted by n n nnn. Each trial is independent of the others, and the number of trials is predetermined.
  2. Two Possible Outcomes:
    • Each trial has exactly two possible outcomes: success (with probability p p ppp) and failure (with probability 1 p 1 p 1-p1 – p1p). These outcomes are mutually exclusive and collectively exhaustive.
  3. Constant Probability of Success:
    • The probability of success, denoted by p p ppp, remains constant for each trial. Similarly, the probability of failure remains 1 p 1 p 1-p1 – p1p throughout the trials.
  4. Independence of Trials:
    • The trials are independent, meaning the outcome of one trial does not affect the outcome of any other trial.
  5. Discrete Distribution:
    • The binomial distribution is discrete, meaning it deals with outcomes that are countable, such as the number of successes in a series of trials.
  6. Probability Mass Function (PMF):
    • The probability of observing exactly k k kkk successes in n n nnn trials is given by the binomial probability mass function:
    P ( X = k ) = ( n k ) p k ( 1 p ) n k P ( X = k ) = ( n k ) p k ( 1 p ) n k P(X=k)=((n)/(k))*p^(k)*(1-p)^(n-k)P(X = k) = \binom{n}{k} \cdot p^k \cdot (1 – p)^{n – k}P(X=k)=(nk)pk(1p)nk
    Where:
    • ( n k ) ( n k ) ((n)/(k))\binom{n}{k}(nk) is the binomial coefficient, representing the number of ways to choose k k kkk successes from n n nnn trials.
    • p p ppp is the probability of success on each trial.
    • 1 p 1 p 1-p1 – p1p is the probability of failure on each trial.
  7. Mean and Variance:
    • The mean μ μ mu\muμ of a binomial distribution is given by:
    μ = n p μ = n p mu=n*p\mu = n \cdot pμ=np
    • The variance σ 2 σ 2 sigma^(2)\sigma^2σ2 of a binomial distribution is given by:
    σ 2 = n p ( 1 p ) σ 2 = n p ( 1 p ) sigma^(2)=n*p*(1-p)\sigma^2 = n \cdot p \cdot (1 – p)σ2=np(1p)
  8. Symmetry and Shape:
    • The shape of the binomial distribution depends on the value of p p ppp:
      • If p = 0.5 p = 0.5 p=0.5p = 0.5p=0.5, the distribution is symmetric.
      • If p < 0.5 p < 0.5 p < 0.5p < 0.5p<0.5, the distribution is skewed to the right (positively skewed).
      • If p > 0.5 p > 0.5 p > 0.5p > 0.5p>0.5, the distribution is skewed to the left (negatively skewed).

Important Uses of Binomial Distribution

The binomial distribution has several important uses in various fields of study. Some of its key applications include:
  1. Quality Control and Manufacturing:
    • The binomial distribution is used to model the number of defective items in a batch or the number of successful outcomes in a series of quality control tests. For example, it can be used to calculate the probability that a certain number of products out of a sample are defective.
  2. Medical Research:
    • In clinical trials, the binomial distribution can be used to model the probability of a certain number of patients responding to a treatment (success) versus not responding (failure). This helps in analyzing the effectiveness of treatments or drugs.
  3. Survey Research and Polling:
    • Binomial distribution is useful in analyzing survey results, where the outcome of interest might be a simple yes/no answer. For example, it can be used to calculate the probability that a certain percentage of respondents will support a particular candidate or policy.
  4. Reliability Testing:
    • In reliability engineering, the binomial distribution can be used to model the probability of a certain number of components in a system failing within a specified period, assuming each component has the same probability of failure.
  5. Genetics:
    • In genetics, the binomial distribution is used to predict the number of offspring with a certain trait in a given number of trials (e.g., the probability that a certain number of offspring will inherit a particular gene).
  6. Marketing and Sales:
    • The binomial distribution is used to model the number of successful sales calls or responses in a given number of attempts, helping to analyze the effectiveness of sales strategies or marketing campaigns.
  7. Finance and Insurance:
    • Binomial distribution can be used in finance to model the number of successful investments or in insurance to model the number of claims filed out of a certain number of policies.

Conclusion

The binomial distribution is a versatile and widely used probability distribution in various fields. Its properties—such as a fixed number of trials, constant probability of success, and independence of trials—make it particularly useful for modeling scenarios with binary outcomes (success or failure). Its applications range from quality control to clinical trials, finance, and marketing, helping researchers and decision-makers analyze and predict the likelihood of specific outcomes.




Question:-04(a)

Fit a straight line Y = a + b X Y = a + b X Y=a+bXY = a + bXY=a+bX to the following data. Compare the estimated values of the dependent variable with its actual values.
X 5 8 10 12 13 15 17 16
Y 8 12 14 10 13 16 14 17
X 5 8 10 12 13 15 17 16 Y 8 12 14 10 13 16 14 17| X | 5 | 8 | 10 | 12 | 13 | 15 | 17 | 16 | | :— | :— | :— | :— | :— | :— | :— | :— | :— | | Y | 8 | 12 | 14 | 10 | 13 | 16 | 14 | 17 |

Answer:

Straight line equation is y = a + b x y = a + b x y=a+bxy=a+b xy=a+bx.
The normal equations are
y = a n + b x x y = a x + b x 2 y = a n + b x x y = a x + b x 2 {:[ sum y=an+b sum x],[ sum xy=a sum x+b sumx^(2)]:}\begin{aligned} & \sum y=a n+b \sum x \\ & \sum x y=a \sum x+b \sum x^2 \end{aligned}y=an+bxxy=ax+bx2
The values are calculated using the following table
x x xxx y y yyy x 2 x 2 x^(2)x^2x2 x y x y x*yx \cdot yxy
5 8 25 40
8 12 64 96
10 14 100 140
12 10 144 120
13 13 169 169
15 16 225 240
17 14 289 238
16 17 256 272
x = 96 x = 96 sum x=96\sum x=96x=96 y = 104 y = 104 sum y=104\sum y=104y=104 x 2 = 1272 x 2 = 1272 sumx^(2)=1272\sum x^2=1272x2=1272 x y = 1315 x y = 1315 sum x*y=1315\sum x \cdot y=1315xy=1315
x y x^(2) x*y 5 8 25 40 8 12 64 96 10 14 100 140 12 10 144 120 13 13 169 169 15 16 225 240 17 14 289 238 16 17 256 272 — — — — sum x=96 sum y=104 sumx^(2)=1272 sum x*y=1315| $x$ | $y$ | $x^2$ | $x \cdot y$ | | :—: | :—: | :—: | :—: | | 5 | 8 | 25 | 40 | | 8 | 12 | 64 | 96 | | 10 | 14 | 100 | 140 | | 12 | 10 | 144 | 120 | | 13 | 13 | 169 | 169 | | 15 | 16 | 225 | 240 | | 17 | 14 | 289 | 238 | | 16 | 17 | 256 | 272 | | — | — | — | — | | $\sum x=96$ | $\sum y=104$ | $\sum x^2=1272$ | $\sum x \cdot y=1315$ |
Substituting these values in the normal equations
8 a + 96 b = 104 96 a + 1272 b = 1315 8 a + 96 b = 104 96 a + 1272 b = 1315 {:[8a+96 b=104],[96 a+1272 b=1315]:}\begin{aligned} & 8 a+96 b=104 \\ & 96 a+1272 b=1315 \end{aligned}8a+96b=10496a+1272b=1315
Solving these two equations using Elimination method,
8 a + 96 b = 104 8 a + 96 b = 104 8a+96 b=1048 a+96 b=1048a+96b=104
8 ( a + 12 b ) = 8 13 a + 12 b = 13 and 96 a + 1272 b = 1315 a + 12 b = 13 ( 1 ) 96 a + 1272 b = 1315 ( 2 ) equation( 1 ) × 96 96 a + 1152 b = 1248 equation ( 2 ) × 1 96 a + 1272 b = 1315 8 ( a + 12 b ) = 8 13 a + 12 b = 13  and  96 a + 1272 b = 1315 a + 12 b = 13 ( 1 ) 96 a + 1272 b = 1315 ( 2 )  equation(  1 ) × 96 96 a + 1152 b = 1248  equation  ( 2 ) × 1 96 a + 1272 b = 1315 {:[8(a+12 b)=8*13],[a+12 b=13],[” and “96 a+1272 b=1315],[a+12 b=13 rarr(1)],[96 a+1272 b=1315 rarr(2)],[” equation( “1)xx96=>96 a+1152 b=1248],[” equation “(2)xx1=>96 a+1272 b=1315]:}\begin{aligned} & 8(a+12 b)=8 \cdot 13 \\ & a+12 b=13 \\ & \text { and } 96 a+1272 b=1315 \\ & a+12 b=13 \rightarrow(1) \\ & 96 a+1272 b=1315 \rightarrow(2) \\ & \text { equation( } 1) \times 96 \Rightarrow 96 a+1152 b=1248 \\ & \text { equation }(2) \times 1 \Rightarrow 96 a+1272 b=1315 \end{aligned}8(a+12b)=813a+12b=13 and 96a+1272b=1315a+12b=13(1)96a+1272b=1315(2) equation( 1)×9696a+1152b=1248 equation (2)×196a+1272b=1315
Substracting 120 b = 67 120 b = 67 =>-120 b=-67\Rightarrow-120 b=-67120b=67
120 b = 67 b = 67 120 b = 0.5583 120 b = 67 b = 67 120 b = 0.5583 {:[=>120 b=67],[=>b=(67)/(120)],[=>b=0.5583]:}\begin{aligned} & \Rightarrow 120 b=67 \\ & \Rightarrow b=\frac{67}{120} \\ & \Rightarrow b=0.5583 \end{aligned}120b=67b=67120b=0.5583
Now substituting this values in the equation is y = a + b x , we get y = 6.3 + 0.5583 x  Now substituting this values in the equation is  y = a + b x , we get  y = 6.3 + 0.5583 x {:[” Now substituting this values in the equation is “y=a+bx”, we get “],[y=6.3+0.5583 x]:}\begin{aligned} &\text { Now substituting this values in the equation is } y=a+b x \text {, we get }\\ &y=6.3+0.5583 x \end{aligned} Now substituting this values in the equation is y=a+bx, we get y=6.3+0.5583x




Question:-04(b)

Define correlation coefficient. What are its properties?

Answer:

Correlation Coefficient

The correlation coefficient is a statistical measure that quantifies the strength and direction of the relationship between two variables. It provides an indication of how much one variable changes in relation to changes in another variable. The most commonly used correlation coefficient is the Pearson correlation coefficient (denoted as r r rrr), which measures the linear relationship between two continuous variables.
The Pearson correlation coefficient is calculated as:
r = Cov ( X , Y ) σ X σ Y r = Cov ( X , Y ) σ X σ Y r=(“Cov”(X,Y))/(sigma _(X)sigma _(Y))r = \frac{\text{Cov}(X, Y)}{\sigma_X \sigma_Y}r=Cov(X,Y)σXσY
Where:
  • Cov ( X , Y ) Cov ( X , Y ) “Cov”(X,Y)\text{Cov}(X, Y)Cov(X,Y) is the covariance between variables X X XXX and Y Y YYY,
  • σ X σ X sigma _(X)\sigma_XσX is the standard deviation of X X XXX,
  • σ Y σ Y sigma _(Y)\sigma_YσY is the standard deviation of Y Y YYY.
The correlation coefficient r r rrr ranges from 1 1 -1-11 to + 1 + 1 +1+1+1, where:
  • r = 1 r = 1 r=1r = 1r=1 indicates a perfect positive linear relationship between the two variables.
  • r = 1 r = 1 r=-1r = -1r=1 indicates a perfect negative linear relationship between the two variables.
  • r = 0 r = 0 r=0r = 0r=0 indicates no linear relationship between the two variables.

Properties of the Correlation Coefficient

The correlation coefficient has several important properties that help in understanding and interpreting the relationship between two variables:
  1. Range of Values:
    • The correlation coefficient r r rrr always lies between 1 1 -1-11 and + 1 + 1 +1+1+1: 1 r 1 1 r 1 -1 <= r <= 1-1 \leq r \leq 11r1
    • A value of r = 1 r = 1 r=1r = 1r=1 indicates a perfect positive linear relationship, r = 1 r = 1 r=-1r = -1r=1 indicates a perfect negative linear relationship, and r = 0 r = 0 r=0r = 0r=0 indicates no linear relationship.
  2. Direction of the Relationship:
    • Positive Correlation: If r > 0 r > 0 r > 0r > 0r>0, this means that as one variable increases, the other variable also tends to increase. The relationship is positively linear.
    • Negative Correlation: If r < 0 r < 0 r < 0r < 0r<0, this means that as one variable increases, the other variable tends to decrease. The relationship is negatively linear.
    • Zero Correlation: If r = 0 r = 0 r=0r = 0r=0, this indicates that there is no linear relationship between the variables.
  3. Symmetry:
    • The correlation coefficient is symmetric. This means the correlation between X X XXX and Y Y YYY is the same as the correlation between Y Y YYY and X X XXX: r ( X , Y ) = r ( Y , X ) r ( X , Y ) = r ( Y , X ) r(X,Y)=r(Y,X)r(X, Y) = r(Y, X)r(X,Y)=r(Y,X)
    • The order of the variables does not affect the value of the correlation coefficient.
  4. Unit-Free Measure:
    • The correlation coefficient is a dimensionless measure, meaning it does not depend on the units of measurement of the variables. Changing the units of the variables (e.g., from inches to centimeters) does not affect the correlation coefficient.
  5. Sensitivity to Linear Relationships:
    • The correlation coefficient only measures the strength of a linear relationship. It may not capture non-linear relationships between variables. Two variables could have a strong non-linear relationship, but their correlation coefficient could be close to zero if the relationship is not linear.
  6. Unaffected by Scale Changes:
    • The correlation coefficient is unaffected by changes in scale (multiplying all values of a variable by a constant) or location (adding a constant to all values of a variable). For example, if X X XXX and Y Y YYY are transformed into X = a X + b X = a X + b X^(‘)=aX+bX’ = aX + bX=aX+b and Y = c Y + d Y = c Y + d Y^(‘)=cY+dY’ = cY + dY=cY+d, the correlation between X X X^(‘)X’X and Y Y Y^(‘)Y’Y will be the same as that between X X XXX and Y Y YYY.
  7. Significance Testing:
    • The correlation coefficient can be tested for statistical significance to determine whether the observed relationship is likely to have occurred by chance. A significance test of the correlation coefficient determines whether the correlation is significantly different from zero in the population.
  8. Effect of Outliers:
    • The correlation coefficient is sensitive to outliers. A single extreme value can significantly distort the value of r r rrr, making it appear stronger or weaker than it actually is.

Interpretation of the Correlation Coefficient

  • Perfect Positive Correlation ( r = 1 r = 1 r=1r = 1r=1): A perfect positive correlation indicates that all data points lie on a straight line with a positive slope, meaning that increases in one variable are associated with proportional increases in the other.
  • Perfect Negative Correlation ( r = 1 r = 1 r=-1r = -1r=1): A perfect negative correlation indicates that all data points lie on a straight line with a negative slope, meaning that increases in one variable are associated with proportional decreases in the other.
  • No Correlation ( r = 0 r = 0 r=0r = 0r=0): No correlation indicates that there is no linear relationship between the variables. However, this does not necessarily mean that there is no relationship at all; there may still be a non-linear relationship.
  • Weak, Moderate, and Strong Correlations:
    • Weak Correlation: If | r | | r | |r||r||r| is between 0 and 0.3, the relationship is considered weak.
    • Moderate Correlation: If | r | | r | |r||r||r| is between 0.3 and 0.7, the relationship is considered moderate.
    • Strong Correlation: If | r | | r | |r||r||r| is between 0.7 and 1, the relationship is considered strong.

Conclusion

The correlation coefficient is a valuable tool for assessing the strength and direction of a linear relationship between two variables. Its key properties—such as symmetry, unit-free nature, and sensitivity to linearity—allow it to be widely applied in various fields, from economics to biology. However, while useful, it has limitations, particularly in capturing non-linear relationships and its susceptibility to outliers. Thus, it should be used with caution and in conjunction with other statistical tools when analyzing data.




Question:-05

What is a life table? Explain its uses and limitations.

Answer:

Life Table

A life table is a statistical tool used to summarize the mortality patterns of a population. It provides a detailed account of the probability of death at various ages, usually for a given cohort (group of individuals born in the same year). The life table is widely used in demography, actuarial science, and public health to assess the longevity and mortality risks of populations or subgroups within a population.
A life table typically includes the following columns:
  1. Age Interval (x): The age group or age interval being analyzed (e.g., 0–1, 1–2, etc.).
  2. Number of Survivors (lx): The number of individuals surviving to a particular age out of a starting population (typically set at 100,000).
  3. Probability of Death (qx): The probability that an individual of age x x xxx will die before reaching the next age.
  4. Number of Deaths (dx): The number of individuals expected to die in the age interval.
  5. Person-Years Lived (Lx): The total number of years lived by the population in the specified age interval.
  6. Life Expectancy (ex): The average number of years remaining for an individual of age x x xxx.
There are two main types of life tables:
  • Cohort Life Table: Follows a group of individuals born in the same year throughout their entire lives, recording their mortality experience as they age.
  • Period Life Table: Reflects mortality conditions of a population at a specific point in time, assuming that current mortality rates remain constant throughout an individual’s life.

Uses of a Life Table

Life tables are widely used in various fields, including demography, epidemiology, public health, actuarial science, and biology. Here are some of the key uses:
  1. Estimating Life Expectancy:
    • Life tables provide estimates of life expectancy at various ages. For example, life expectancy at birth is a commonly used demographic indicator that summarizes the average number of years a newborn is expected to live given current mortality rates.
  2. Evaluating Mortality Risks:
    • Life tables give detailed information on the probability of death at different ages, allowing for the evaluation of mortality risks. This is particularly useful for actuarial calculations, such as determining insurance premiums or pension contributions.
  3. Public Health and Policy:
    • Life tables can be used to assess health trends and the effectiveness of health interventions. They help policymakers understand population dynamics and plan for healthcare resources by identifying the ages where mortality risks are highest.
  4. Actuarial Science and Insurance:
    • Actuaries use life tables to calculate insurance premiums, pension liabilities, and other long-term financial products based on the mortality risks of individuals or populations. Life tables provide a foundation for estimating the risk of death and life expectancy, which are crucial for pricing life insurance and annuities.
  5. Population Forecasting:
    • Life tables are used in population projections to predict future population sizes, age distributions, and demographic changes. By adjusting for fertility, migration, and mortality rates, life tables contribute to accurate population modeling.
  6. Comparative Studies:
    • Life tables allow for the comparison of mortality rates across different populations, regions, or time periods. For example, comparing life tables of different countries can reveal disparities in health and life expectancy due to varying socioeconomic and healthcare conditions.
  7. Evaluating the Impact of Diseases or Epidemics:
    • Life tables can be modified to assess the impact of diseases (such as HIV/AIDS or COVID-19) on a population’s mortality patterns and life expectancy, helping to quantify the effects of epidemics or public health crises.

Limitations of a Life Table

While life tables are valuable tools, they do have several limitations:
  1. Assumption of Constant Mortality Rates:
    • Life tables often assume that mortality rates remain constant over time, especially in period life tables. In reality, mortality rates may change due to medical advancements, lifestyle changes, or socioeconomic shifts, which can make projections inaccurate if the rates are not adjusted accordingly.
  2. Cohort vs. Period Life Tables:
    • Cohort life tables require data over a long period of time, as they track a cohort from birth to death. This makes them impractical for current mortality assessments. Period life tables, on the other hand, provide a snapshot based on current mortality rates but do not account for future changes, potentially underestimating or overestimating life expectancy.
  3. Exclusion of Migration:
    • Life tables typically do not account for migration, focusing solely on mortality. Migration can significantly alter population dynamics, especially in regions with high immigration or emigration rates.
  4. No Insight into Cause-Specific Mortality:
    • Life tables do not provide information on the causes of death. They summarize overall mortality patterns but do not differentiate between deaths caused by different diseases, accidents, or other factors.
  5. Homogeneity Assumption:
    • Life tables generally assume that all individuals in a given age group have the same mortality risk. This assumption of homogeneity ignores variations in mortality risk due to factors such as gender, socioeconomic status, ethnicity, or access to healthcare.
  6. Data Quality:
    • The accuracy of a life table depends on the quality of the data used to construct it. Incomplete or inaccurate mortality data can lead to errors in estimating life expectancy and mortality probabilities.
  7. Exclusion of Morbidity:
    • Life tables focus only on mortality and do not account for morbidity (illness or disability). While they can estimate the likelihood of death at different ages, they do not provide information about the quality of life or the number of years lived in good health.

Conclusion

A life table is a powerful statistical tool used to analyze the mortality and survival patterns of a population, estimate life expectancy, and evaluate mortality risks. It plays a critical role in public health, actuarial science, and demographic studies. However, life tables also have limitations, such as assuming constant mortality rates and excluding factors like migration and cause-specific mortality. Despite these limitations, life tables remain essential for understanding population dynamics and planning for future demographic trends.




Assignment III

Question:-06

Write short notes on the following:
(a) Bayes’ theorem of probability
(b) Age specific birth and death rates
(c) Measurement of Skewness

Answer:

(a) Bayes’ Theorem of Probability

Bayes’ Theorem is a fundamental result in probability theory that describes how to update the probability of a hypothesis based on new evidence. It provides a way to revise existing predictions or theories in light of new information. The formula for Bayes’ Theorem is:
P ( A | B ) = P ( B | A ) P ( A ) P ( B ) P ( A | B ) = P ( B | A ) P ( A ) P ( B ) P(A|B)=(P(B|A)*P(A))/(P(B))P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)}P(A|B)=P(B|A)P(A)P(B)
Where:
  • P ( A | B ) P ( A | B ) P(A|B)P(A|B)P(A|B) is the posterior probability, or the probability of event A A AAA occurring given that B B BBB has occurred.
  • P ( B | A ) P ( B | A ) P(B|A)P(B|A)P(B|A) is the likelihood, or the probability of event B B BBB occurring given that A A AAA has occurred.
  • P ( A ) P ( A ) P(A)P(A)P(A) is the prior probability of event A A AAA, representing the initial belief before observing B B BBB.
  • P ( B ) P ( B ) P(B)P(B)P(B) is the marginal likelihood, or the total probability of event B B BBB.
Use: Bayes’ Theorem is widely used in various fields, such as statistics, machine learning, medicine, and decision-making under uncertainty. It helps update the probability of a hypothesis based on new data, for example, in diagnosing diseases or in spam email filtering.

(b) Age-Specific Birth and Death Rates

Age-Specific Birth Rate:
  • Definition: The age-specific birth rate (ASBR) measures the number of live births per 1,000 women of a specific age group in a given year. It is often used to analyze fertility patterns across different age groups in a population.
  • Formula: A S B R = Number of Births to Women in Age Group Number of Women in Age Group × 1000 A S B R = Number of Births to Women in Age Group Number of Women in Age Group × 1000 ASBR=(“Number of Births to Women in Age Group”)/(“Number of Women in Age Group”)xx1000ASBR = \frac{\text{Number of Births to Women in Age Group}}{\text{Number of Women in Age Group}} \times 1000ASBR=Number of Births to Women in Age GroupNumber of Women in Age Group×1000
  • Example: ASBR could be calculated for women aged 20–24, 25–29, etc., to understand at what ages women are most likely to have children.
Age-Specific Death Rate:
  • Definition: The age-specific death rate (ASDR) measures the number of deaths per 1,000 individuals of a specific age group in a given year. It provides insight into mortality patterns across different ages.
  • Formula: A S D R = Number of Deaths in Age Group Population in Age Group × 1000 A S D R = Number of Deaths in Age Group Population in Age Group × 1000 ASDR=(“Number of Deaths in Age Group”)/(“Population in Age Group”)xx1000ASDR = \frac{\text{Number of Deaths in Age Group}}{\text{Population in Age Group}} \times 1000ASDR=Number of Deaths in Age GroupPopulation in Age Group×1000
  • Example: ASDR can be calculated for children (0–4 years), young adults (15–24 years), or older adults (65+ years) to assess the mortality risk in different life stages.
Use: Both ASBR and ASDR are crucial in demographic analysis, helping public health officials and policymakers design targeted interventions for specific age groups, such as maternal health programs or geriatric care.

(c) Measurement of Skewness

Skewness is a measure of the asymmetry or departure from symmetry of a probability distribution of a random variable. It helps determine whether the data is concentrated on one side of the distribution.
  1. Positive Skewness (Right-Skewed):
    • The distribution has a long right tail, meaning that the right side (higher values) extends farther than the left. Most data points are concentrated on the left.
    • Example: Income distributions often exhibit positive skewness because a few individuals have significantly higher incomes than the rest.
  2. Negative Skewness (Left-Skewed):
    • The distribution has a long left tail, meaning that the left side (lower values) extends farther than the right. Most data points are concentrated on the right.
    • Example: The distribution of exam scores where most students do well but a few score very low.
  3. Formula: Skewness is calculated using the following formula:
    Skewness = n ( n 1 ) ( n 2 ) ( x i x ¯ s ) 3 Skewness = n ( n 1 ) ( n 2 ) x i x ¯ s 3 “Skewness”=(n)/((n-1)(n-2))sum((x_(i)-( bar(x)))/(s))^(3)\text{Skewness} = \frac{n}{(n-1)(n-2)} \sum \left( \frac{x_i – \bar{x}}{s} \right)^3Skewness=n(n1)(n2)(xix¯s)3
    Where:
    • n n nnn is the number of observations,
    • x i x i x_(i)x_ixi is each individual observation,
    • x ¯ x ¯ bar(x)\bar{x}x¯ is the mean of the data,
    • s s sss is the standard deviation of the data.
  4. Interpretation:
    • Skewness = 0: The distribution is symmetric.
    • Skewness > 0: The distribution is positively skewed (right-skewed).
    • Skewness < 0: The distribution is negatively skewed (left-skewed).
Use: Skewness is used in statistics to assess the shape of data distributions, aiding in the selection of appropriate statistical tests and models. It is particularly important in financial modeling, risk assessment, and quality control where understanding the symmetry of data distributions can inform decision-making.




Question:-07

Differentiate between the following:
(a) Simple random sampling and Stratified random sampling
(b) Type I and Type II errors in hypothesis testing
(c) Estimator and Estimate

Answer:

(a) Simple Random Sampling vs. Stratified Random Sampling

Simple Random Sampling:
  • Definition: Simple random sampling is a method in which every member of the population has an equal chance of being selected in the sample. It is the most basic form of probability sampling.
  • Procedure: A simple random sample is typically drawn using methods like a random number generator or drawing names from a hat. Each member is selected independently of the others.
  • Example: Selecting 50 students at random from a school of 500 students, where each student has an equal chance of being chosen.
  • Advantages:
    • Easy to implement and understand.
    • Reduces bias, as every member has an equal chance of being included.
  • Disadvantages:
    • May not be representative if the population is heterogeneous (i.e., has subgroups with differing characteristics).
Stratified Random Sampling:
  • Definition: Stratified random sampling involves dividing the population into strata (subgroups) based on certain characteristics (e.g., age, income, education level), and then drawing a random sample from each stratum.
  • Procedure: The population is first divided into mutually exclusive strata, and then a simple random sample is taken from each stratum, either proportionally or equally.
  • Example: Dividing a school’s students into grades (strata) and then randomly selecting a certain number of students from each grade for the sample.
  • Advantages:
    • Increases the representativeness of the sample, especially in heterogeneous populations.
    • Ensures that specific subgroups are adequately represented in the sample.
  • Disadvantages:
    • More complex and time-consuming than simple random sampling.
    • Requires detailed knowledge of the population structure.

(b) Type I and Type II Errors in Hypothesis Testing

Type I Error:
  • Definition: A Type I error occurs when the null hypothesis H 0 H 0 H_(0)H_0H0 is rejected when it is actually true. It is also known as a false positive.
  • Probability: The probability of committing a Type I error is denoted by α α alpha\alphaα, which is the significance level of the test (commonly set at 0.05 or 0.01).
  • Example: Concluding that a new drug is effective when, in reality, it is not.
  • Consequence: A Type I error leads to the acceptance of an incorrect alternative hypothesis, potentially resulting in false conclusions and incorrect actions.
Type II Error:
  • Definition: A Type II error occurs when the null hypothesis H 0 H 0 H_(0)H_0H0 is not rejected when it is actually false. It is also known as a false negative.
  • Probability: The probability of committing a Type II error is denoted by β β beta\betaβ, and the power of a test is 1 β 1 β 1-beta1 – \beta1β, representing the probability of correctly rejecting a false null hypothesis.
  • Example: Concluding that a new drug is not effective when, in reality, it is.
  • Consequence: A Type II error leads to the failure to detect an effect or difference when one actually exists, potentially causing missed opportunities for beneficial outcomes.

Summary of Errors:

  • Type I Error: Rejecting H 0 H 0 H_(0)H_0H0 when it is true (false positive).
  • Type II Error: Failing to reject H 0 H 0 H_(0)H_0H0 when it is false (false negative).

(c) Estimator vs. Estimate

Estimator:
  • Definition: An estimator is a statistical rule or formula used to calculate an estimate of a population parameter based on sample data. It is a function or method that provides the mechanism for estimation.
  • Properties: Estimators are evaluated based on properties like unbiasedness, efficiency, consistency, and sufficiency.
  • Example: The sample mean X ¯ X ¯ bar(X)\bar{X}X¯ is an estimator of the population mean μ μ mu\muμ; the sample variance S 2 S 2 S^(2)S^2S2 is an estimator of the population variance σ 2 σ 2 sigma^(2)\sigma^2σ2.
Estimate:
  • Definition: An estimate is the actual numerical value obtained by applying the estimator to a sample of data. It is the specific value calculated as a result of using the estimator on the sample.
  • Example: If you calculate the average height of 50 randomly selected students from a school and find it to be 5.6 feet, then 5.6 feet is the estimate of the population mean height μ μ mu\muμ.

Summary of Estimator vs. Estimate:

  • Estimator: A rule or method used to derive an estimate (e.g., sample mean formula).
  • Estimate: The actual calculated value based on the sample data (e.g., the average height calculated).




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