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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)  0  10  20  30  40  50 
No. of Students  50  46  40  20  10  03 
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 Students  f  c.f 
0—10  5  50  50 – 46 =4  0 + 4 = 4 
10—20  15  46  46 – 40 =6  4 + 6 = 10 
20—30  25  40  40 – 20 =20  10 + 20 = 30 
30—40  35  20  20 – 10 = 10  30 + 10 = 40 
40—50  45  10  10 – 3 =7  40 + 7 = 47 
50—60  55  03  3 – 0 = 3  47 + 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
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
20 falls in c.f of 30 so so l=20, h = 10, f=20, 40n/100 = 20, c=10
Formula
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(BA) = 0.85
The Probability of unsuccessful products have received favourable survey indications =P(B¬A)= 0.30
Bayes Theorem
First we have to find:
P(B) = P(BA) x P(A) + P(B¬A) x P(¬A)
P(B) = (0.85)(0.65) + (0.30)(0.35)P(B) = 0.6575
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.)  
Commodity  2013  2015  2013  2015 
Rise  100  90  9.3  4.5 
Wheat  11  10  6.4  3.7 
Pulses  5  3  5.1  2.7 
(3 Marks)
Solution:
Dorbish –Bowley’s Method is also called LP method
Formula
Items  2013  2015  
Price (Po)  Quantity (qo)  Price (P1)  Quantity (q1)  poqo  p1qo  p1q1  poq1  
A  9.3  100  4.5  90  930  450  405  837 
B  6.4  11  3.7  10  70.4  40.7  37  64 
C  5.1  5  2.7  3  25.5  13.5  8.1  15.3 
D  
Sum  1025.9  504.2  450.1  916.3  
∑poqo =  ∑p1qo =  ∑p1q1 =  ∑poq1 = 
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 Per  Company A  Company B  
Employee  Male  Female  Male  Female 
Mean  30  20  27  32 
Variance  8  3  12  5 
No of Employees  40  10  20  30 
(6 Marks)
Solution:
Data Company A:
Data Company B:
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:
Year  2010  2011  2012  2013  2014  2015 
Salary  10420  10432  10960  11300  11900  12500 
CPI  126.8  129.5  136.2  141.2  152.3  165.4 
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:
Year  Salary (1)  CPI (2) 

2010  10420  126.8  8217.67 
2011  10432  129.5  8055.60 
2012  10960  136.2  8046.99 
2013  11300  141.2  8002.83 
2014  11900  152.3  7813.53 
2015  12500  165.4  7557.44 
To maintain the standard in 2015 as equal to 2011, his salary in 2015 should be:
(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:
 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.
 Second axiom is that the probability of entire sample is 1.
 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 nonnegative 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:
Year  2009  2010  2011  2012  2013  2014  2015 
Production of Steel (m.t)  80  90  92  83  94  99  92 
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:
Year  Production of Steel (m.t) (y)  x = t – 2012  x²  xy  Trend Values y = 90 + 2x 
2009  80  3  9  240  90 + 2(3)=84 
2010  90  2  4  180  90 + 2(2)=86 
2011  92  1  1  92  90 + 2(1)=88 
2012  83  0  0  0  90 + 2(0)=90 
2013  94  1  1  94  90 + 2(1)=92 
2014  99  2  4  198  90 + 2(2)=94 
2015  92  3  9  276  90 + 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
y = a + bx
y = 90 + 2x
Elimination of trend using multiplicative model
Year  Production of Steel (m.t) (y)  ye  Multiplicative Model =(y/ye) 
2009  80  84  80/84=0.9523 
2010  90  86  90/86=1.0465 
2011  92  88  92/88=1.0454 
2012  83  90  83/90=0.9222 
2013  94  92  94/92=1.0217 
2014  99  94  99/94=1.0531 
2015  92  96  92/96=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 fastfood 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 fastfood 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 onehour 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
Step 4: Critical Region
10.05 = 0.95, 0.95/2 = 0.475 falls in z table 1.9 column and sixth row so 1.96
Z>1.96
Step 5: Calculation
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.
Year  Motor Registration  No. of Tyres Sold 
1  600  1250 
2  630  1100 
3  720  1300 
4  750  1350 
5  800  1500 
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:
Year  Motor Registration (x)  No. of Tyres Sold (y)  xy  x²  y² 
1  600  1250  750000  360000  1562500 
2  630  1100  693000  396900  1210000 
3  720  1300  936000  518400  1690000 
4  750  1350  1012500  562500  1822500 
5  800  1500  1200000  640000  2250000 
Sum  3500  6500  4591500  2477800  8535000 
∑x =  ∑y =  ∑xy =  ∑x² =  ∑y² = 
Line of Regression Y on X
Where:
Estimation of sale of tyres when registration is 850
(b) The final exam in a oneterm 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:
(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?
(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)
20x – 9y – 107 = 0….(ii)
Check
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)
20x – 9y – 107 = 0….(ii)
Check
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:
Regression line Y on X
Regression line X on Y
Now we have to find σy where σx=3
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?
Standard normal random variable is greater than the value of z, so:
Go to Z Normal Probability Table and find 0.4 row and 3^{rd} column, we have a value 0.6664
(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 4^{th} & 5^{th} column.
So:
Formula Applied
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:  2  4  5  6  8  11 
Y:  18  12  10  8  7  5 
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
x  y  Dx=(xx̅)  Dy=(yy̅)  DxDy  Dx²  Dy² 
2  18  4  8  32  16  64 
4  12  2  2  4  4  4 
5  10  1  0  0  1  0 
6  8  0  2  0  0  4 
8  7  2  3  6  4  9 
11  5  5  5  25  25  25 
36  60  67  50  106  
∑x=  ∑y=  ∑DxDy=  ∑Dx²=  ∑Dy²= 
Condition 2
Formula
x  y  U=2x+6  V=3y15  UV  U²  V² 
2  18  10  39  390  100  1521 
4  12  14  21  294  196  441 
5  10  16  15  240  256  225 
6  8  18  9  162  324  81 
8  7  22  6  132  484  36 
11  5  28  0  0  784  0 
108  90  1218  2144  2304  
∑U=  ∑V=  ∑UV=  ∑U²=  ∑V²= 
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 onetailed and twotailed tests with reference to testing of hypothesis.
(3 Marks)
Answer:
A twotailed or two sided test
A twotailed or two sided test also known as a nondirectional hypothesis, is the standard test of significance to determine if there is a relationship between variables in either direction. Twotailed tests do this by dividing the .05 in two and putting half on each side of the bell curve.
A onetailed or one sided test
A onetailed 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.
Business  2.86  2.77  3.18  2.80  3.14  2.87  3.19  3.24  2.91  3.00  2.83  
Arts & Science  3.35  3.32  3.36  3.63  3.41  3.37  3.45  3.43  3.44  3.17  3.26  3.18  3.41 
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
Where
Step 4: Critical Region
Where v = n1 + n2 – 2 = 11 + 13 – 2 =22
Step 5: Calculation
Business (X1)  Arts & Science (X2)  X1²  X2² 
2.86  3.35  8.1796  11.2225 
2.77  3.32  7.6729  11.0224 
3.18  3.36  10.1124  11.2896 
2.8  3.63  7.84  13.1769 
3.14  3.41  9.8596  11.6281 
2.87  3.37  8.2369  11.3569 
3.19  3.45  10.1761  11.9025 
3.24  3.43  10.4976  11.7649 
2.91  3.44  8.4681  11.8336 
3  3.17  9  10.0489 
2.83  3.26  8.0089  10.6276 
3.18  10.1124  
3.41  11.6281  
32.79  43.78  98.0521  147.6144 
∑X1=  ∑X2=  ∑ X1² =  ∑ X2² = 
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
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 beboth 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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