Solved Paper Statistics for Business Decisions 2016 BMS CBCS Delhi University

Solved Paper Statistics for Business Decisions 2016 BMS CBCS Delhi University
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In this post, I am going to present you the Solved Paper Statistics for Business Decisions 2016 BMS CBCS Delhi University for BMS Bachelor of Management Studies. Stay in touch for other study notes and solutions of the papers related to Statistics, Accounting, Finance, economics & other resource of Investment, Corporate finance and business.

Solved by Iftikhar Ali Lecturer Business & Financial Studies

Table of Contents

Solved Paper Statistics for Business Decisions 2016 BMS CBCS Delhi University

Q.1: (a) Following is the distribution of marks in Statistics obtained by 50 students. Calculate the median marks. What does it represent? If 60% of the students pass this test, find the minimum marks obtained by a pass candidate.

Q.1: (a) Following is the distribution of marks in Statistics obtained by 50 students.

Marks (More than)01020304050
No. of Students504640201003

Calculate the medianmarks. What does it represent? If 60% of the students pass this test, find the minimum marks obtained by a pass candidate. (6 Marks)

Solution:

Marks (More than)Class Mark (X)No. of Studentsfc.f
0—1055050 – 46 =40  + 4 = 4
10—20154646 – 40 =64 + 6 = 10
20—30254040 – 20 =2010 + 20 = 30
30—40352020 – 10 = 1030 + 10 = 40
40—50451010 – 3 =740 + 7 = 47
50—6055033 – 0 = 347 + 3 = 50
   ∑f = n = 50 

Selection of model class for median

n/2 = 50/2 =25 falls in c.f of 30 so l=20, h = 10, f=20, n/2 = 25, c=10

Formula

    \[ \mathbf{Median = l +}\frac{\mathbf{h}}{\mathbf{f}}\left( \frac{\mathbf{n}}{\mathbf{2}}\mathbf{- c} \right)\ \]

    \[ \mathbf{Median = 20 +}\frac{\mathbf{10}}{\mathbf{20}}\left( \mathbf{25 - 10} \right)\ \]

    \[ \mathbf{Median = 27.5}\ \]

Median 27.5 indicates that average marks obtained by students are 27.5

60% of 50 students = 50 x 0.60 = 30 students

Since 60% of the students pass the test. It means that 40% students get less than minimum pass marks. So we have to calculate Percentile 40 (P40) in order to know the minimum marks.

Selection of model class for P40

    \[ \frac{\mathbf{40}\mathbf{n}}{\mathbf{100}}\mathbf{=}\frac{\mathbf{40 \times 50}}{\mathbf{100}}\mathbf{=}\frac{\mathbf{2000}}{\mathbf{100}}\mathbf{= 20}\ \]

20 falls in c.f of 30 so so l=20, h = 10, f=20, 40n/100 = 20, c=10

Formula

    \[ \mathbf{P}_{\mathbf{40}}\mathbf{= l +}\frac{\mathbf{h}}{\mathbf{f}}\left( \frac{\mathbf{40}\mathbf{n}}{\mathbf{100}}\mathbf{- c} \right)\ \]

    \[ \mathbf{P}_{\mathbf{40}}\mathbf{= 20 +}\frac{\mathbf{10}}{\mathbf{20}}\left( \mathbf{20 - 10} \right)\ \]

    \[ \mathbf{P}_{\mathbf{40}}\mathbf{= 25}\ \]

Minimum marks obtained by pass students are 25

(b) An organisation dealing with consumer products wants to introduce a new product in the market. Based on past experience, it has 65 percent chance of being successful……….85 percent of the time, whereas unsuccessful products have received favourable survey indications 30 percent of the time. Determine the probability of the product being successful given that the survey information is favourable.

(b) An organisation dealing with consumer products wants to introduce a new product in the market. Based on past experience, it has 65 percent chance of being successful. In order to help the organisation to make a decision on the new product that is, whether to introduce the product or not, it decides to get additional information on consumer attitude towards the product. For this, the organisation decides to conduct a survey. In the past, when a product of this type was successful, the survey had yielded favourable indications 85 percent of the time, whereas unsuccessful products have received favourable survey indications 30 percent of the time. Determine the probability of the product being successful given that the survey information is favourable. (6 Marks)

Solution:

Here we can use Bayes Theorem

Let’s Denote:

Event that product is successful = A

Event that survey is indicating Successful = B

Given that:

The prior probability that the product is successful = P(A) = 0.65

The prior probability that the product is not successful = P(¬A) = 0.35

The probability of survey indicating that the product is successful = P(B|A)  = 0.85

The Probability of unsuccessful products have received favourable survey indications =P(B|¬A)= 0.30

Bayes Theorem

    \[ \mathbf{P}\left( \mathbf{A} \middle| \mathbf{B} \right)\mathbf{=}\frac{\mathbf{P(B|A) \times P(A)}}{\mathbf{P(B)}}\ \]

First we have to find:

P(B) = P(B|A)  x P(A) + P(B|¬A) x P(¬A)

P(B) = (0.85)(0.65) + (0.30)(0.35)P(B) = 0.6575

    \[ \mathbf{P}\left( \mathbf{A} \middle| \mathbf{B} \right)\mathbf{=}\frac{\mathbf{P(B|A) \times P(A)}}{\mathbf{P(B)}}\ \]

    \[ \mathbf{P}\left( \mathbf{A} \middle| \mathbf{B} \right)\mathbf{=}\frac{\mathbf{0.85\ \times 0.65}}{\mathbf{0.6575}}\ \]

    \[ \mathbf{P}\left( \mathbf{A} \middle| \mathbf{B} \right)\mathbf{= 0.84}\ \]

So the probability of the product being successful as per the favorable survey is 84%

(c) Calculate the Price index using the Dorbish and Bowley’s method for the following data with 2013 as the base year.

(c) Calculate the Price index using the Dorbish and Bowley’s method for the following data with 2013 as the base year:

 Quantity (in “000” Kg)Price (in Rs.)
Commodity2013201520132015
Rise100909.34.5
Wheat11106.43.7
Pulses535.12.7

(3 Marks)

Solution:

Dorbish –Bowley’s Method is also called LP method

Formula

    \[ \mathbf{P}_{\mathbf{n}}\mathbf{DB =}\frac{\mathbf{1}}{\mathbf{2}}\left\lbrack \frac{\mathbf{\sum p}\mathbf{1}\mathbf{q}\mathbf{0}}{\mathbf{\sum p}\mathbf{0}\mathbf{q}\mathbf{0}}\mathbf{+}\frac{\mathbf{\sum p}\mathbf{1}\mathbf{q}\mathbf{1}}{\mathbf{\sum p}\mathbf{0}\mathbf{q}\mathbf{1}}\mathbf{\ } \right\rbrack\mathbf{\times 100}\ \]

  Items20132015 
Price (Po)Quantity (qo)Price   (P1)Quantity (q1)poqop1qop1q1poq1
A9.31004.590930450405837
B6.4113.71070.440.73764
C5.152.7325.513.58.115.3
D        
Sum1025.9504.2450.1916.3
 ∑poqo =∑p1qo =∑p1q1 =∑poq1 =

    \[ \mathbf{P}_{\mathbf{2015}}\mathbf{DB =}\frac{\mathbf{1}}{\mathbf{2}}\left\lbrack \frac{\mathbf{\sum p}\mathbf{1}\mathbf{q}\mathbf{0}}{\mathbf{\sum p}\mathbf{0}\mathbf{q}\mathbf{0}}\mathbf{+}\frac{\mathbf{\sum p}\mathbf{1}\mathbf{q}\mathbf{1}}{\mathbf{\sum p}\mathbf{0}\mathbf{q}\mathbf{1}}\mathbf{\ } \right\rbrack\mathbf{\times 100}\ \]

    \[ \mathbf{P}_{\mathbf{2015}}\mathbf{DB =}\frac{\mathbf{1}}{\mathbf{2}}\left\lbrack \frac{\mathbf{504.2}}{\mathbf{1025.9}}\mathbf{+}\frac{\mathbf{450.1}}{\mathbf{916.3}}\mathbf{\ } \right\rbrack\mathbf{\times 100}\ \]

    \[ \mathbf{P}_{\mathbf{2015}}\mathbf{DB =}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{\ (0.4914 + 0.4912) \times 100}\ \]

    \[ \mathbf{P}_{\mathbf{2015}}\mathbf{DB = 49.13}\ \]

Q.2: (a) The following data is given for two companies. Combining the data for groups of male and female employees, (i) Find out which company has a higher average productivity per employee, and(ii) Which company has more consistent productivity?

Q.2: (a) The following data is given for two companies. Combining the data for groups of male and female employees,

(i) Find out which company has a higher average productivity per employee, and

(ii) Which company has more consistent productivity?

Productivity PerCompany ACompany B
EmployeeMaleFemaleMaleFemale
Mean30202732
Variance83125
No of Employees40102030

(6 Marks)

Solution:

Data Company A:

    \[ \overline{\mathbf{X}}\mathbf{m = 30,\ }\overline{\mathbf{X}}\mathbf{f = 20} \]

    \[ \mathbf{S}^{\mathbf{2}}\mathbf{m = 8,\ }\mathbf{S}^{\mathbf{2}}\mathbf{f = 3} \]

    \[ \mathbf{n}{\mathbf{m}}\mathbf{= 40,\ }\mathbf{n}{\mathbf{f}}\mathbf{= 10} \]

    \[ {\overline{\mathbf{X}}}{\mathbf{c}}\mathbf{=}\frac{\mathbf{n}{\mathbf{m}}\overline{\mathbf{X}}\mathbf{m +}\mathbf{n}{\mathbf{f}}\overline{\mathbf{X}}\mathbf{f}}{\mathbf{n}{\mathbf{m}}\mathbf{+}\mathbf{n}_{\mathbf{f}}}\ \]

    \[ {\overline{\mathbf{X}}}_{\mathbf{c}}\mathbf{=}\frac{\mathbf{(40 \times 30) + (10 \times 20)}}{\mathbf{40 + 10}}\ \]

    \[ {\overline{\mathbf{X}}}_{\mathbf{c}}\mathbf{=}\frac{\mathbf{1200 + 200}}{\mathbf{50}}\ \]

    \[ {\overline{\mathbf{X}}}_{\mathbf{c}}\mathbf{= 28}\ \]

    \[ \mathbf{Combined\ S.D =}\sqrt{\frac{\mathbf{n}{\mathbf{m}}\left\lbrack \mathbf{S}^{\mathbf{2}}\mathbf{m +}{\mathbf{(}\overline{\mathbf{X}}\mathbf{m -}{\overline{\mathbf{X}}}{\mathbf{c}}\mathbf{)}}^{\mathbf{2}} \right\rbrack\mathbf{+}\mathbf{n}{\mathbf{f}}\left\lbrack \mathbf{S}^{\mathbf{2}}\mathbf{f +}{\mathbf{(}\overline{\mathbf{X}}\mathbf{f -}{\overline{\mathbf{X}}}{\mathbf{c}}\mathbf{)}}^{\mathbf{2}} \right\rbrack\mathbf{\ }}{\mathbf{n}{\mathbf{m}}\mathbf{+}\mathbf{n}{\mathbf{f}}}}\ \]

    \[ \mathbf{Combined\ S.D =}\sqrt{\frac{\mathbf{40}\left\lbrack \mathbf{8 +}\mathbf{(30 - 28)}^{\mathbf{2}} \right\rbrack\mathbf{+ 10}\left\lbrack \mathbf{3 +}\mathbf{(20 - 28)}^{\mathbf{2}} \right\rbrack\mathbf{\ }}{\mathbf{40 + 10}}}\ \]

    \[ \mathbf{Combined\ S.D =}\sqrt{\frac{\mathbf{48 + 670\ }}{\mathbf{50}}}\ \]

    \[ \mathbf{Combined\ S.D = 3.789}\ \]

    \[ \mathbf{Combined\ C.V =}\frac{\mathbf{Combined\ S.D}}{\mathbf{Combined\ Mean}}\mathbf{\times 100}\ \]

    \[ \mathbf{Combined\ C.V =}\frac{\mathbf{3.789}}{\mathbf{28}}\mathbf{\times 100 = 13.53\%}\ \]

Data Company B:

    \[ \overline{\mathbf{X}}\mathbf{m = 27,\ }\overline{\mathbf{X}}\mathbf{f = 32} \]

    \[ \mathbf{S}^{\mathbf{2}}\mathbf{m = 12,\ }\mathbf{S}^{\mathbf{2}}\mathbf{f = 5} \]

    \[ \mathbf{n}{\mathbf{m}}\mathbf{= 20,\ }\mathbf{n}{\mathbf{f}}\mathbf{= 30}\ \]

    \[ {\overline{\mathbf{X}}}{\mathbf{c}}\mathbf{=}\frac{\mathbf{n}{\mathbf{m}}\overline{\mathbf{X}}\mathbf{m +}\mathbf{n}{\mathbf{f}}\overline{\mathbf{X}}\mathbf{f}}{\mathbf{n}{\mathbf{m}}\mathbf{+}\mathbf{n}_{\mathbf{f}}}\ \]

    \[ {\overline{\mathbf{X}}}_{\mathbf{c}}\mathbf{=}\frac{\mathbf{(20 \times 27) + (30 \times 32)}}{\mathbf{20 + 30}}\ \]

    \[ {\overline{\mathbf{X}}}_{\mathbf{c}}\mathbf{=}\frac{\mathbf{540 + 960}}{\mathbf{50}}\ \]

    \[ {\overline{\mathbf{X}}}_{\mathbf{c}}\mathbf{= 30}\ \]

    \[ \mathbf{Combined\ S.D =}\sqrt{\frac{\mathbf{n}{\mathbf{m}}\left\lbrack \mathbf{S}^{\mathbf{2}}\mathbf{m +}{\mathbf{(}\overline{\mathbf{X}}\mathbf{m -}{\overline{\mathbf{X}}}{\mathbf{c}}\mathbf{)}}^{\mathbf{2}} \right\rbrack\mathbf{+}\mathbf{n}{\mathbf{f}}\left\lbrack \mathbf{S}^{\mathbf{2}}\mathbf{f +}{\mathbf{(}\overline{\mathbf{X}}\mathbf{f -}{\overline{\mathbf{X}}}{\mathbf{c}}\mathbf{)}}^{\mathbf{2}} \right\rbrack\mathbf{\ }}{\mathbf{n}{\mathbf{m}}\mathbf{+}\mathbf{n}{\mathbf{f}}}}\ \]

    \[ \mathbf{Combined\ S.D =}\sqrt{\frac{\mathbf{20}\left\lbrack \mathbf{12 +}\mathbf{(27 - 30)}^{\mathbf{2}} \right\rbrack\mathbf{+ 30}\left\lbrack \mathbf{5 +}\mathbf{(32 - 30)}^{\mathbf{2}} \right\rbrack\mathbf{\ }}{\mathbf{20 + 30}}}\ \]

    \[ \mathbf{Combined\ S.D =}\sqrt{\frac{\mathbf{420 + 270\ }}{\mathbf{50}}}\ \]

    \[ \mathbf{Combined\ S.D = 3.71}\ \]

    \[ \mathbf{Combined\ C.V =}\frac{\mathbf{Combined\ S.D}}{\mathbf{Combined\ Mean}}\mathbf{\times 100}\ \]

    \[ \mathbf{Combined\ C.V =}\frac{\mathbf{3.71}}{\mathbf{30}}\mathbf{\times 100 = 12.36\%}\ \]

Results:

Company A Combine Mean = 28

Company A Combined C.V = 13.53%

Company B Combine Mean = 30

Company B Combined C.V = 12.36%

Conclusion:

Company B has Higher Productivity since Combine Mean of Company B is greater than Company A. 30 > 28

Company B is more consistent in productivity since combined C.V of Company B is less than the Combined C.V of Company A. 12.36 < 13.53

(b) The employees of an organisation have presented the following data in support of their contention that they are entitled to a wage increase. The data represents average monthly Salary of the employees……..Compute the Real Wages for the period 2010 to 2015. Find the amount of Salary required in the year 2015 to ensure the purchasing power equal to that enjoyed in 2011.

(b) The employees of an organisation have presented the following data in support of their contention that they are entitled to a wage increase. The data represents average monthly Salary of the employees:

Year201020112012201320142015
Salary104201043210960113001190012500
CPI126.8129.5136.2141.2152.3165.4

Compute the Real Wages for the period 2010 to 2015. Find the amount of Salary required in the year 2015 to ensure the purchasing power equal to that enjoyed in 2011. (6 Marks)

Solution:

YearSalary (1)CPI (2)

    \[ \mathbf{Real\ Monthly\ Salary =}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{\ \times 100}\ \]

201010420126.88217.67
201110432129.58055.60
201210960136.28046.99
201311300141.28002.83
201411900152.37813.53
201512500165.47557.44

To maintain the standard in 2015 as equal to 2011, his salary in 2015 should be:

    \[ \frac{165.4}{100} \times 10432 = 17255\ \]

(c) What is Probability? Discuss the three approaches for calculating the probability with the help of appropriate examples.

Answer:

Probability is a statistical way or method to calculate the chance of happening or occurring some event or not happening or not occurring some event. Sum of both events is equals to 1. There are three approaches of probability such as:

  • Classical Approach
  • Empirical Approach
  • Axiomatic Approach

Classical Approach

Classical Approach is a simple approach of probability in which every event has equal chance to occur or happen.

For example in throwing a die each event e.g., 1, 2, 3, 4, 5, 6 has equal chance to appear 1/6. Similarly in throwing a coin, both head and tail have equal chance ½.

Empirical Approach

Empirical approach of probability is based on observed data which is observed in experiments. For example you roll a six sided fair cubical die 100 time and you noted that 5 appears 30 times in 100 experiments so you have empirical evidence so probability of getting 5 will be 30/100 = 0.30

Axiomatic Approach

Under axiomatic approach of probability, we have to follow some rules or axioms of probability. These rules or axioms are formulated by Kolmogorov. These axioms are listed below:

  1. First axiom is that the least possible probability is 0 and maximum possible probability is 1. First axiom is also called non negative axiom means it cannot be negative.
  2. Second axiom is that the probability of entire sample is 1.
  3. Third axiom is that two mutually exclusive events cannot come together.

For example probability of drawing spade out of 52 playing cards is:

P(Spade) = Total Spades/Total Cards = 13/52=0.25

Above example satisfies all three axioms such that probability is non-negative 0.25, second, if all 52 cards selected as event then 52/52 = 1 and third each event is mutually exclusive.

Q.3: (a) A Company that manufactures Steel observed the production of steel (in metric tonnes) as follows………..Fit a straight line trend to the given data using the method of least squares. Calculate the trend values and eliminate trend using the Multiplication model. Estimate the production of steel for the year 20l7.

Q.3: (a) A Company that manufactures Steel observed the production of steel (in metric tonnes) as follows:

Year2009201020112012201320142015
Production of Steel (m.t)80909283949992

Fit a straight line trend to the given data using the method of least squares. Calculate the trend values and eliminate trend using the Multiplication model. Estimate the production of steel for the year 20l7.

(9 Marks)

Solution:

YearProduction of Steel (m.t) (y)x = t – 2012  x²xyTrend Values y = 90 + 2x  
200980-39-24090 + 2(-3)=84
201090-24-18090 + 2(-2)=86
201192-11-9290 + 2(-1)=88
20128300090 + 2(0)=90
201394119490 + 2(1)=92
2014992419890 + 2(2)=94
2015923927690 + 2(3)=96
n=7∑y = 630∑x = 0∑x² = 28∑xy = 56∑ye=627

Fitting straight Line

Let the straight line:

y = a + bx where 

    \[ \mathbf{a =}\frac{\mathbf{\sum y}}{\mathbf{n}}\mathbf{,\ b =}\frac{\mathbf{\sum xy}}{\mathbf{\sum}\mathbf{x}^{\mathbf{2}}}\ \]

    \[ \mathbf{a =}\frac{\mathbf{\sum y}}{\mathbf{n}}\mathbf{=}\frac{\mathbf{630}}{\mathbf{7}}\mathbf{= 90}\ \]

    \[ \mathbf{b =}\frac{\mathbf{\sum xy}}{\mathbf{\sum}\mathbf{x}^{\mathbf{2}}}\mathbf{=}\frac{\mathbf{56}}{\mathbf{28}}\mathbf{= 2}\ \]

y = a + bx

y = 90 + 2x

Elimination of trend using multiplicative model

YearProduction of Steel (m.t) (y)yeMultiplicative Model =(y/ye)
2009808480/84=0.9523
2010908690/86=1.0465
2011928892/88=1.0454
2012839083/90=0.9222
2013949294/92=1.0217
2014999499/94=1.0531
2015929692/96=0.9583

Estimate the production of steel for the year 20l7

First x = t – 2012 = 2017 – 2012 = 5

Now putting x = 5 into straight line equation:

y = 90 + 2x

y = 90 + 2(5)

y = 100 so the production of steel for the year 2017 will be 100 m.t.

(b) The manager of a fast-food restaurant has to determine whether the population mean waiting time to place an order has changed in the past month from its previous population mean value of 4.5 minutes. From past experience, it is assumed that the population is normally distributed, with a standard deviation of 1.2 minutes…..

(b) The manager of a fast-food restaurant has to determine whether the population mean waiting time to place an order has changed in the past month from its previous population mean value of 4.5 minutes. From past experience, it is assumed that the population is normally distributed, with a standard deviation of 1.2 minutes. A sample of 25 orders during a one-hour period is selected. The sample mean is 5.1 minutes. Determine whether there is evidence at the 0.05 level of significance that the population mean waiting time to place an order has changed in the past month from its previous population mean value of 4.5 minutes. (6 Marks)

Solution:

Step 1: Stating Null and Alternative Hypothesis

Ho: Null Hypotheses = µ=4.5

H1: Alternative Hypotheses = µ ≠ 4.5

Step 2: Level of significance

Alpha α = 0.05, n = 25

Step 3: Test Statistics

    \[ \mathbf{Z =}\frac{\overline{\mathbf{x}}\mathbf{-}\mathbf{\mu}_{\mathbf{0}}}{\frac{\mathbf{\sigma}}{\sqrt{\mathbf{n}}}}\ \]

Step 4: Critical Region

1-0.05 = 0.95, 0.95/2 = 0.475 falls in z table 1.9 column and sixth row so 1.96

Z>1.96

    \[ \mathbf{Z}{\mathbf{\alpha}}\mathbf{=}\mathbf{Z}{\mathbf{0.05}}\mathbf{= \pm 1.96)}\ \]

Step 5: Calculation

    \[ \mathbf{Z =}\frac{\overline{\mathbf{x}}\mathbf{-}\mathbf{\mu}_{\mathbf{0}}}{\frac{\mathbf{\sigma}}{\sqrt{\mathbf{n}}}}\ \]

    \[ \mathbf{Z =}\frac{\mathbf{5.1 - 4.5}}{\frac{\mathbf{1.2}}{\sqrt{\mathbf{25}}}}\mathbf{=}\frac{\mathbf{0.6}}{\mathbf{0.24}}\mathbf{= 2.5}\ \]

Step 6: Conclusion

Since calculated value of z statistic 2.5 exceeds than the tabulated value 1.96 or it falls in critical region so we reject null hypothesis and accept alternative hypothesis and we can say that the mean time to place an order has changed.

Q.4: (a) The table given below shows the number of motor registrations in a certain city for a term of five years and the sale of motor tyres by a firm in that city for the same period…..Find the regression equation to estimate the sale of tyres when motor registration is known. Estimate sale of tyres when registration is 850.

Q.4: (a) The table given below shows the number of motor registrations in a certain city for a term of five years and the sale of motor tyres by a firm in that city for the same period.

YearMotor RegistrationNo. of Tyres Sold
16001250
26301100
37201300
47501350
58001500

Find the regression equation to estimate the sale of tyres when motor registration is known. Estimate sale of tyres when registration is 850.                                               (6 Marks)

Solution:

YearMotor Registration (x)No. of Tyres Sold (y)xyx²y²
160012507500003600001562500
263011006930003969001210000
372013009360005184001690000
4750135010125005625001822500
5800150012000006400002250000
Sum35006500459150024778008535000
 ∑x =∑y =∑xy =∑x² =∑y² =

Line of Regression Y on X

    \[ \widehat{\mathbf{Y}}\mathbf{= a + bx}\ \]

Where:

    \[ \mathbf{a =}\overline{\mathbf{Y}}\mathbf{- b}\overline{\mathbf{X}}\mathbf{\ \ \&\ }\mathbf{b}_{\mathbf{yx}}\mathbf{=}\frac{\mathbf{\sum xy - n}\overline{\mathbf{X}}\overline{\mathbf{Y}}}{\mathbf{\sum}\mathbf{x}^{\mathbf{2}}\mathbf{- n}{\mathbf{(}\overline{\mathbf{X}}\mathbf{)}}^{\mathbf{2}}}\mathbf{\ \ \ }\ \]

    \[ \overline{\mathbf{X}}\mathbf{=}\frac{\mathbf{\sum x}}{\mathbf{n}}\mathbf{=}\frac{\mathbf{3500}}{\mathbf{5}}\mathbf{= 700}\ \]

    \[ \overline{\mathbf{Y}}\mathbf{=}\frac{\mathbf{\sum y}}{\mathbf{n}}\mathbf{=}\frac{\mathbf{6500}}{\mathbf{5}}\mathbf{= 1300}\ \]

    \[ \mathbf{b}_{\mathbf{yx}}\mathbf{=}\frac{\mathbf{\sum xy - n}\overline{\mathbf{X}}\overline{\mathbf{Y}}}{\mathbf{\sum}\mathbf{x}^{\mathbf{2}}\mathbf{- n}{\mathbf{(}\overline{\mathbf{X}}\mathbf{)}}^{\mathbf{2}}}\mathbf{\ \ \ }\ \]

    \[ \mathbf{b}_{\mathbf{yx}}\mathbf{=}\frac{\mathbf{4591500 - 5(700)(1300)}}{\mathbf{2477800}\mathbf{- 5}\mathbf{(700)}^{\mathbf{2}}}\mathbf{\ \ \ }\ \]

    \[ \mathbf{b}_{\mathbf{yx}}\mathbf{=}\frac{\mathbf{41500}}{\mathbf{27800}}\mathbf{= 1.4928\ \ }\ \]

    \[ \mathbf{a =}\overline{\mathbf{Y}}\mathbf{- b}\overline{\mathbf{X}}\ \]

    \[ \mathbf{a = 1300 - (1.4928)(700)} \]

    \[ \mathbf{a = 255.04} \]

    \[ \widehat{\mathbf{Y}}\mathbf{= a + bx} \]

    \[ \widehat{\mathbf{Y}}\mathbf{= 255.04 + 1.4928}\mathbf{x} \]

Estimation of sale of tyres when registration is 850

    \[ \widehat{\mathbf{Y}}\mathbf{= 255.04 + 1.4928}\mathbf{x}\ \]

    \[ \widehat{\mathbf{Y}}\mathbf{= 255.04 + 1.4928(850)}\ \]

    \[ \widehat{\mathbf{Y}}\mathbf{= 1524\ tyres\ Approx}\ \]

(b) The final exam in a one-term statistics course is taken in the December exam period. Students who are sick or have other legitimate reasons for missing the exam are allowed to write a deferred exam scheduled for the first week in January. A Statistics professor has observed that only 2% of all students legitimately miss the December final exam. Suppose that the professor has 40 students registered this term.

(i) How many students can the professor expect to miss the December exam?

(ii) What is the probability that the professor will not have to create a deferred exam?

(6 Marks)

Solution:

Data:

    \[ \mathbf{n = 40,\ p = 0.02,\ q = 1 - p = 1 - 0.02 = 0.98}\ \]

(i) How many students can the professor expect to miss the December exam?

    \[ \mathbf{Mean\ E}\left( \mathbf{X} \right)\mathbf{= np = 40 \times 0.02 = 0.8\ }\ \]

(ii) What is the probability that the professor will not have to create a deferred exam?

    \[ \mathbf{P}\left\lbrack \mathbf{X = x} \right\rbrack\mathbf{=}\frac{\mathbf{n!}}{\mathbf{x!}\left( \mathbf{n - 0} \right)\mathbf{!}}\mathbf{\times}\mathbf{p}^{\mathbf{x}}\mathbf{\times}\mathbf{q}^{\mathbf{n - x}}\ \]

    \[ \mathbf{P}\left\lbrack \mathbf{X = 0} \right\rbrack\mathbf{=}\frac{\mathbf{40!}}{\mathbf{0!}\left( \mathbf{40 - 0} \right)\mathbf{!}}\left( \mathbf{0.02} \right)^{\mathbf{0}}\mathbf{(0.98)}^{\mathbf{40 - 0}}\mathbf{= 0.4457}\ \]

(c) If the two lines of regression are 4x – 5y + 30 = 0 and 20x – 9y – 107 = 0. Which of these is the line of regression of X on Y and Y on X. Find rxy, and σy when σx=3.

(3 Marks)

Solution:

Let the first equation be the line of regression x on y & the second be the y on x

4x – 5y + 30 = 0…..(i)

    \[ 4x = 5y - 30\ \ \]

    \[ x = \frac{5y}{4} - \frac{30}{4}\ldots.here\ b_{xy} = \frac{5}{4}\ \]

20x – 9y – 107 = 0….(ii)

    \[ -\ 9y\ = \ - 20x\ + 107\ \ \ \]

    \[ 9y\ = \ 20x - 107\ \ \ \]

    \[ y\ = \frac{20x}{9} - \frac{107}{9}\ldots.\ here\ b_{yx} = \frac{20}{9}\ \]

Check

    \[ \mathbf{r}_{\mathbf{xy}}\mathbf{=}\sqrt{\mathbf{bxy \times byx}}\ \]

    \[ \mathbf{r}_{\mathbf{xy}}\mathbf{=}\sqrt{\frac{\mathbf{5}}{\mathbf{4}}\mathbf{\times}\frac{\mathbf{20}}{\mathbf{9}}}\mathbf{=}\sqrt{\mathbf{2.78}}\mathbf{= 1.67}\ \]

Above answer does not satisfy the property of Correlation Coefficient -1≤ r ≤+1 so our selection of equations for regression line is wrong.

Let the first equation be the line of regression y on x & the second be the x on y

4x – 5y + 30 = 0…..(i)

    \[ - 5y = - 4x - 30\ \]

    \[ 5y = 4x + 30\ \]

    \[ y = \frac{4x}{5} + \frac{30}{5}\ldots.here\ b_{yx} = \frac{4}{5} \]

20x – 9y – 107 = 0….(ii)

    \[ 20x\ = \ 9y\ + 107\ \ \ \]

    \[ x\ = \frac{9y}{20} + \frac{107}{20}\ldots.\ here\ b_{xy} = \frac{9}{20}\ \]

Check

    \[ \mathbf{r}_{\mathbf{xy}}\mathbf{=}\sqrt{\mathbf{bxy \times byx}}\ \]

    \[ \mathbf{r}_{\mathbf{xy}}\mathbf{=}\sqrt{\frac{\mathbf{9}}{\mathbf{20}}\mathbf{\times}\frac{\mathbf{4}}{\mathbf{5}}}\mathbf{=}\sqrt{\mathbf{0.36}}\mathbf{=}\mathbf{0.6}\ \]

Above answer satisfies the property of Correlation Coefficient -1≤ r ≤+1 so our selection of equations for regression line is correct and following results have obtained:

    \[ \mathbf{r}{\mathbf{xy}}\mathbf{= 0.6,\ }b{yx} = \frac{4}{5},\ b_{xy} = \frac{9}{20}\ \]

Regression line Y on X

    \[ \widehat{\mathbf{y}}\mathbf{= a + bx} \]

    \[ \widehat{\mathbf{y}}\mathbf{=}\frac{\mathbf{30}}{\mathbf{5}}\mathbf{+}\frac{\mathbf{4}}{\mathbf{5}}\mathbf{x} \]

    \[ \widehat{\mathbf{y}}\mathbf{=}\mathbf{6}\mathbf{+}\mathbf{0.8}\mathbf{x} \]

Regression line X on Y

    \[ \widehat{\mathbf{x}}\mathbf{= a + b}\mathbf{y} \]

    \[ \widehat{\mathbf{x}}\mathbf{=}\frac{\mathbf{107}}{\mathbf{20}}\mathbf{+}\frac{\mathbf{9}}{\mathbf{20}}\mathbf{y} \]

    \[ \widehat{\mathbf{x}}\mathbf{=}\mathbf{5.35}\mathbf{+}\mathbf{0.45}\mathbf{y} \]

Now we have to find σy where σx=3

    \[ b_{yx} = \mathbf{r}_{\mathbf{xy}}\left( \frac{\mathbf{\sigma y}}{\mathbf{\sigma x}} \right)\mathbf{\ }\ \]

    \[ 0.8 = \mathbf{0.6}\left( \frac{\mathbf{\sigma}\mathbf{y}}{\mathbf{3}} \right)\ \]

    \[ \left( \frac{0.8}{0.6} \right)3 = \mathbf{\sigma y}\ \]

    \[ \mathbf{\sigma y}\mathbf{= 4}\ \]

Q.5: (a) Because of the relatively high interest rates, most consumers attempt to pay off their credit card bills promptly. However, this is not always possible. An analysis of the amount of interest paid monthly by a bank’s Visa cardholders reveals that the amount is normally distributed with a mean of 27 Dollars and a standard deviation of 7 Dollars.

(i) What proportion of the bank’s Visa card holders pay more than 30 Dollars in interest?(ii) What interest payment is exceeded by only 20% of the bank’s Visa cardholders?    

 (6 Marks)

Solution:

(i) What proportion of the bank’s Visa card holders pay more than $30 in interest?

    \[ P(X>30)=P\left(\frac{X-Mu}\sigma>\frac{30-27}7\right) \]

    \[ \mathbf{P}\left( \mathbf{X > 30} \right)\mathbf{= P(Z > 0.43)}\ \]

Standard normal random variable is greater than the value of z, so:

    \[ \mathbf{P}\left( \mathbf{X > 30} \right)\mathbf{= 1 - P(Z < 0.43)}\ \]

Go to Z Normal Probability Table and find 0.4 row and 3rd column, we have a value 0.6664

    \[ \mathbf{P}\left( \mathbf{X > 30} \right)\mathbf{= 1 - 0.6664}\ \]

    \[ \mathbf{P}\left( \mathbf{X > 30} \right)\mathbf{= 0.3336}\ \]

(ii) What interest payment is exceeded by only 20% of the bank’s Visa cardholders?    

First we go to the Z Normal Probability Table and locate 1 – 0.20 =0.80 that will be found in 0.8 row and between 4th & 5th column.

So:

    \[ Z_{0.20} = 0.845,\ µ = 27,\ \sigma = 7,\ X = ?\ \]

Formula Applied

    \[ \mathbf{Z}_{\mathbf{0.20}}\mathbf{=}\frac{\mathbf{X - Mu}}{\mathbf{\sigma}} \]

    \[ \mathbf{0.845 =}\frac{\mathbf{X - 27}}{\mathbf{7}} \]

    \[ \mathbf{0.845 \times 7 = X - 27} \]

    \[ \mathbf{X = 5.915 + 27} \]

    \[ \mathbf{X = 32.915} \]

Interest amount of 32.915 is exceeded by 20%

(b) Compute Karl Pearson’s correlation coefficient between the corresponding values of X and Y from the following table…….Now suppose each value of variable X is multiplied by 2 and 6 is added to the product. Further, each value of Y is multiplied with 3 and 15 is subtracted from the product. Comment whether correlation coefficient will change or not.

(b) Compute Karl Pearson’s correlation coefficient between the corresponding values of X and Y from the following table:

X:2456811
Y:181210875

Now suppose each value of variable X is multiplied by 2 and 6 is added to the product. Further, each value of Y is multiplied with 3 and 15 is subtracted from the product. Comment whether correlation coefficient will change or not.                  (6 Marks)

Solution:

Condition 1

Formula

    \[ \mathbf{r =}\frac{\mathbf{\sum(x -}\overline{\mathbf{x}}\mathbf{)(y -}\overline{\mathbf{y}}\mathbf{)}}{\sqrt{\mathbf{\sum}\left( \mathbf{x -}\overline{\mathbf{x}} \right)^{\mathbf{2}}\mathbf{\sum}\left( \mathbf{y -}\overline{\mathbf{y}} \right)^{\mathbf{2}}}}\mathbf{\ or}\ \]

    \[ \mathbf{r =}\frac{\mathbf{\sum DxDy}}{\sqrt{\mathbf{\sum}\mathbf{Dx}^{\mathbf{2}}\mathbf{\sum}\mathbf{Dy}^{\mathbf{2}}}}\ \]

xyDx=(x-x̅)Dy=(y-y̅)DxDyDx²Dy²
218-48-321664
412-22-444
510-10010
680-2004
872-3-649
11555-252525
3660-6750106
∑x=∑y=  ∑DxDy=∑Dx²=∑Dy²=

    \[ \overline{\mathbf{x}}\mathbf{=}\frac{\mathbf{\sum x}}{\mathbf{n}}\mathbf{=}\frac{\mathbf{36}}{\mathbf{6}}\mathbf{= 6}\ \]

    \[ \overline{\mathbf{y}}\mathbf{=}\frac{\mathbf{\sum y}}{\mathbf{n}}\mathbf{=}\frac{\mathbf{60}}{\mathbf{6}}\mathbf{= 10}\ \]

    \[ \mathbf{r =}\frac{\mathbf{\sum DxDy}}{\sqrt{\mathbf{\sum}\mathbf{Dx}^{\mathbf{2}}\mathbf{\sum}\mathbf{Dy}^{\mathbf{2}}}}\ \]

    \[ \mathbf{r =}\frac{\mathbf{- 67}}{\sqrt{\mathbf{50 \times 106}}}\mathbf{= - 0.92}\ \]

Condition 2

Formula

    \[ \mathbf{r}_{\mathbf{xy}}\mathbf{=}\frac{\mathbf{n\sum UV -}\left( \mathbf{\sum U} \right)\left( \mathbf{\sum V} \right)}{\sqrt{\mathbf{n\sum}\mathbf{U}^{\mathbf{2}}\mathbf{- (\sum}\mathbf{U)}^{\mathbf{2}}}\sqrt{\mathbf{n\sum}\mathbf{V}^{\mathbf{2}}\mathbf{- (\sum}\mathbf{V)}^{\mathbf{2}}}}\ \]

xyU=2x+6V=3y-15UVU²V²
21810393901001521
4121421294196441
5101615240256225
6818916232481
8722613248436
11528007840
  10890121821442304
  ∑U=  ∑V=∑UV=∑U²=∑V²=

    \[ \mathbf{r}_{\mathbf{xy}}\mathbf{=}\frac{\mathbf{n\sum UV -}\left( \mathbf{\sum U} \right)\left( \mathbf{\sum V} \right)}{\sqrt{\mathbf{n\sum}\mathbf{U}^{\mathbf{2}}\mathbf{- (\sum}\mathbf{U)}^{\mathbf{2}}}\sqrt{\mathbf{n\sum}\mathbf{V}^{\mathbf{2}}\mathbf{- (\sum}\mathbf{V)}^{\mathbf{2}}}}\ \]

    \[ \mathbf{r}_{\mathbf{xy}}\mathbf{=}\frac{\mathbf{6 \times 1218 -}\left( \mathbf{108} \right)\left( \mathbf{90} \right)}{\sqrt{\mathbf{6 \times 2144 - (108}\mathbf{)}^{\mathbf{2}}}\sqrt{\mathbf{6 \times 2304 - (90}\mathbf{)}^{\mathbf{2}}}}\ \]

    \[ \mathbf{r}_{\mathbf{xy}}\mathbf{=}\frac{\mathbf{7308 - 9720}}{\mathbf{34.64 \times 75.65}}\ \]

    \[ \mathbf{r}_{\mathbf{xy}}\mathbf{=}\frac{\mathbf{- 2412}}{\mathbf{2621}}\ \]

    \[ \mathbf{r}_{\mathbf{xy}}\mathbf{= - 0.92}\ \]

Result: both variables X and Y are highly negatively correlated and both methods produced same results because correlation coefficient is independent to origin and scale.

(c) Differentiate between one-tailed and two-tailed tests with reference to testing of hypothesis.

(3 Marks)

Answer:

A two-tailed or two sided test

A two-tailed or two sided test also known as a non-directional hypothesis, is the standard test of significance to determine if there is a relationship between variables in either direction. Two-tailed tests do this by dividing the .05 in two and putting half on each side of the bell curve.

A one-tailed or one sided test

A one-tailed or one sided test also known as a directional hypothesis, is a test of significance to determine if there is a relationship between the variables in one direction.

Q.6: (a) The Dean ofstudents at a College is wondering about grade distributions at the school. She has heard grumblings that the GPAs in the Business School are about 0.25 lower than those in the College of Arts and Science. A quick random sampling produced the following GPAs.

Q.6: (a) The Dean ofstudents at a College is wondering about grade distributions at the school. She has heard grumblings that the GPAs in the Business School are about 0.25 lower than those in the College of Arts and Science. A quick random sampling produced the following GPAs.

Business2.862.773.182.803.142.873.193.242.913.002.83  
Arts & Science3.353.323.363.633.413.373.453.433.443.173.263.183.41

Does the above data indicate that there is a factual basis for the grumblings? State and test appropriate hypothesis at alpha = 0.05?                    (6 Marks)

Solution:

Step 1: Hypothesis

Null Hypothesis Ho: There is no difference in the mean GPAs between the Business School and the College of Arts and Sciences µ Business = µ Arts & Science

Alternative Hypothesis H1: Business GPA is lower than Arts & Science GPA µ Business < µ Arts & Science

Step 2: Level of Significance

Alpha α = 0.05

Step 3: Test Statistics

    \[ \mathbf{t =}\frac{\mathbf{(x\overline{}}{\mathbf{1}}\mathbf{-}{\overline{\mathbf{x}}}{\mathbf{2)}}\mathbf{-}\mathbf{(\mu}{\mathbf{1}}\mathbf{-}\mathbf{\mu}{\mathbf{2}}\mathbf{)}}{\mathbf{sp}\sqrt{\left( \frac{\mathbf{1}}{\mathbf{n}{\mathbf{1}}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{n}{\mathbf{2}}} \right)}}\ \]

Where

    \[ \mathbf{Sp}\mathbf{=}\sqrt{\frac{\left( \mathbf{n}\mathbf{1 - 1} \right)\mathbf{S}\mathbf{1² +}\left( \mathbf{n}\mathbf{2 - 1} \right)\mathbf{S}\mathbf{2²}}{\mathbf{n}\mathbf{1 + n}\mathbf{2 - 2}}}\ \]

Step 4: Critical Region

    \[ \mathbf{t}{\mathbf{\alpha\ }\left( \mathbf{v} \right)}\mathbf{=}\mathbf{t}{\mathbf{0.05}\mathbf{\ }\left( \mathbf{22} \right)}\mathbf{=}\mathbf{- 1.717}\ \]

Where v = n1 + n2 – 2 = 11 + 13 – 2 =22

Step 5: Calculation

Business (X1)Arts & Science (X2)X1²X2²
2.863.358.179611.2225
2.773.327.672911.0224
3.183.3610.112411.2896
2.83.637.8413.1769
3.143.419.859611.6281
2.873.378.236911.3569
3.193.4510.176111.9025
3.243.4310.497611.7649
2.913.448.468111.8336
33.17910.0489
2.833.268.008910.6276
 3.1810.1124
 3.4111.6281
32.7943.7898.0521147.6144
∑X1=∑X2=∑ X1² =∑ X2² =

    \[ \overline{\mathbf{X}}\mathbf{1 =}\frac{\mathbf{\sum X}\mathbf{1}}{\mathbf{n}\mathbf{1}}\mathbf{=}\frac{\mathbf{32.79}}{\mathbf{11}}\mathbf{=}\mathbf{2.9}\mathbf{8}\ \]

    \[ \overline{\mathbf{X}}\mathbf{2}\mathbf{=}\frac{\mathbf{\sum X}\mathbf{2}}{\mathbf{n}\mathbf{2}}\mathbf{=}\frac{\mathbf{43.78}}{\mathbf{1}\mathbf{2}}\mathbf{=}\mathbf{3.3676}\ \]

    \[ \mathbf{S}\mathbf{1}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{n}\mathbf{1 - 1}}\left\lbrack \mathbf{\sum X}\mathbf{1}^{\mathbf{2}}\mathbf{-}\frac{\left( \mathbf{\sum X}\mathbf{1} \right)^{\mathbf{2}}}{\mathbf{n}\mathbf{1}} \right\rbrack\ \]

    \[ \mathbf{S}\mathbf{1}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{11}\mathbf{- 1}}\left\lbrack \mathbf{98.0521}\mathbf{-}\frac{\left( \mathbf{32.79} \right)^{\mathbf{2}}}{\mathbf{11}} \right\rbrack\ \]

    \[ \mathbf{S}\mathbf{1}^{\mathbf{2}}\mathbf{=}\mathbf{0.03080}\ \]

    \[ \mathbf{S}\mathbf{2}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{n}\mathbf{2}\mathbf{- 1}}\left\lbrack \mathbf{\sum X}\mathbf{2}^{\mathbf{2}}\mathbf{-}\frac{\left( \mathbf{\sum X}\mathbf{2} \right)^{\mathbf{2}}}{\mathbf{n}\mathbf{1}} \right\rbrack\ \]

    \[ \mathbf{S}\mathbf{2}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{1}\mathbf{3}\mathbf{- 1}}\left\lbrack \mathbf{147.6144}\mathbf{-}\frac{\left( \mathbf{43.78} \right)^{\mathbf{2}}}{\mathbf{1}\mathbf{3}} \right\rbrack\ \]

    \[ \mathbf{S}\mathbf{2}^{\mathbf{2}}\mathbf{= 0.0}\mathbf{1473}\ \]

    \[ \mathbf{Sp =}\sqrt{\frac{\left( \mathbf{n}\mathbf{1 - 1} \right)\mathbf{S}\mathbf{1² +}\left( \mathbf{n}\mathbf{2 - 1} \right)\mathbf{S}\mathbf{2²}}{\mathbf{n}\mathbf{1 + n}\mathbf{2 - 2}}}\ \]

    \[ \mathbf{Sp =}\sqrt{\frac{\left( \mathbf{11}\mathbf{- 1} \right)\mathbf{0.03080}\mathbf{+}\left( \mathbf{13}\mathbf{- 1} \right)\mathbf{0.01473}}{\mathbf{11}\mathbf{+}\mathbf{13}\mathbf{- 2}}}\ \]

    \[ \mathbf{Sp =}\sqrt{\frac{\mathbf{0.48476}}{\mathbf{22}}}\mathbf{= 0.1484}\ \]

    \[ \mathbf{t =}\frac{\mathbf{(x\overline{}}{\mathbf{1}}\mathbf{-}{\overline{\mathbf{x}}}{\mathbf{2)}}\mathbf{-}\mathbf{(\mu}{\mathbf{1}}\mathbf{-}\mathbf{\mu}{\mathbf{2}}\mathbf{)}}{\mathbf{sp}\sqrt{\left( \frac{\mathbf{1}}{\mathbf{n}{\mathbf{1}}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{n}{\mathbf{2}}} \right)}}\ \]

    \[ \mathbf{t =}\frac{\mathbf{(2.98 - 3.3676)}\mathbf{-}\mathbf{(0}\mathbf{)}}{\mathbf{0}\mathbf{.1484}\sqrt{\left( \frac{\mathbf{1}}{\mathbf{11}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{13}} \right)}}\ \]

    \[ \mathbf{t =}\frac{\mathbf{- 0.3876}}{\mathbf{0.06079}}\ \]

    \[ \mathbf{t =}\mathbf{- 6.3760}\ \]

Step 6: Conclusion

Our calculated value – 6.3760 is much less than tabulated value – 1.717 or we can say that our calculated value falls in critical or rejection region so we reject null hypothesis and accept alternative hypothesis.

(b) The following information was obtained from the records of a factory relating to the wages: Arithmetic Mean: Rs. 56.80; Median: Rs. 59.50; Standard Deviation: Rs. 12.40 Calculate Mode, Coefficient of Variation and Coefficient of Skewness. Also, comment on the type of distribution.     

(5 Marks)

Solution:

Mode = 3Median – 2Mean

Mode = 3(59.50) – 2(56.80)

Mode = 178.5 – 113.6

Mode = 64.9

    \[ \mathbf{C.V = \ }\frac{\mathbf{S.D}}{\mathbf{Mean}}\mathbf{\times 100} \]

    \[ \mathbf{C.V = \ }\frac{\mathbf{12.40}}{\mathbf{56.80}}\mathbf{\times 100} \]

    \[ \mathbf{C.V =}\mathbf{21.83} \]

    \[ \mathbf{Karl\ Pearso}\mathbf{n}^{\mathbf{'}}\mathbf{s\ C.S =}\frac{\mathbf{Mean - Mode}}{\mathbf{S.D}} \]

    \[ \mathbf{Karl\ Pearso}\mathbf{n}^{\mathbf{'}}\mathbf{s\ C.S =}\frac{\mathbf{56.80}\mathbf{-}\mathbf{64.9}}{\mathbf{12.40}} \]

    \[ \mathbf{Karl\ Pearso}\mathbf{n}^{\mathbf{'}}\mathbf{s\ C.S =}\mathbf{\ - 0.65} \]

Distribution is negatively skewed in which mean is less than mode and distribution is skewed to left with extent of 0.65.

(c) Explain the terms independent and mutually exclusive events. When will the events A and B be-both independent and mutually exclusive?      

(4 Marks)

Answer:

Independent Event

Independent events are events where the occurrence of one event does not affect the probability of the occurrence of another event. Formally, events A and B are independent if and only if:

P(A∩B) = P(A) x P(B)

For example we have two coins. On both coins let’s say A and B, the probability of getting head is ½ for each, so probability of both events will be ½ x ½ = ¼ since each flip is independent to other.

Mutually Exclusive Events

If two or more than two events such as event A and event B have nothing common between them, events are called mutually exclusive events. For example we have an outcomes of throwing die 1, 2, 3, 4, 5, 6 and have two events such as event A an even outcome and B an odd outcome:

All possible outcomes: 1, 2, 3, 4, 5, 6

Even outcome A={2, 4, 6}, Odd outcome B ={1, 3, 5}

So both events will be considered as mutually exclusive events because they both have nothing common between them.

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