By QB365 on 31 Dec, 2022
QB365 provides a detailed and simple solution for every Possible Questions in Class 12 Business Maths Subject - Important 3 Mark English Medium. It will help Students to get more practice questions, Students can Practice these question papers in addition to score best marks.
12th Standard
Business Maths and Statistics
Answer all the following Questions.
If A=\(\left( \begin{matrix} 1 & 1 & -1 \\ 2 & -3 & 4 \\ 3 & -2 & 3 \end{matrix} \right) \) and B=\(\left( \begin{matrix} 1 & -2 & 3 \\ -2 & 4 & -6 \\ 5 & 1 & -1 \end{matrix} \right) \), then find the rank of AB and the rank of BA.
At marina two types of games viz., Horse riding and Quad Bikes riding are available on hourly rent. Keren and Benita spent Rs. 780 and Rs. 560 during the month of May.
Name | Number of hours | Total amount spent (in Rs) |
|
Horse Riding | Quad Bike Riding | ||
Keren | 3 | 4 | 780 |
Benita | 2 | 3 | 560 |
Find the hourly charges for the two games (rides). (Use determinant method).
Evaluate \(\int { \frac { 7x-1 }{ { x }^{ 2 }-5x+6 } dx } \)
Evaluate \(\int { \left( { x }^{ 2 }-2x+5 \right) } { e }^{ -x }dx\)
The cost of over haul of an engine is Rs. 10,000 The operating cost per hour is at the rate of 2x − 240 where the engine has run x km. Find out the total cost if the engine run for 300 hours after overhaul.
For the marginal revenue function MR = 6 − 3x2 − x3, Find the revenue function and demand function.
Solve \(y d x-x d y-3 x^{2} y^{2} e^{x^{3}} d x=0\)
Solve: \(\frac { 1+{ x }^{ 2 } }{ 1+y } =xy\frac { dy }{ dx } \)
From the following table find the missing value
x | 2 | 3 | 4 | 5 | 6 |
f(x) | 45.0 | 49.2 | 54.1 | - | 67.4 |
The following data relates to indirect labour expenses and the level of output
Months | Jan | Feb | Mar | Apr | May | June |
Units of output | 200 | 300 | 400 | 640 | 540 | 580 |
Indirect labour expenses (Rs) | 2500 | 2800 | 3100 | 3820 | 3220 | 3640 |
Estimate the expenses at a level of output of 350 units, by using graphic method
Two unbiased dice are thrown simultaneously and sum of the upturned faces considered as random variable. Construct a probability mass function.
If f (x) is defined by f(x)=ke-2x, 0\(\le\)x<\(\infty\) is a density function. Determine the constant k and also find mean.
When counting red blood cells, a square grid is used, over which a drop of blood is evenly distributed. Under the microscope an average of 8 erythrocytes are observed per single square. What is the probability that exactly 5 erythrocytes are found in one square?
The mortality rate for a certain disease is 7 in 1000. What is the probability for just 2 deaths on account of this disease in a group of 400? [Given e(–2.8) = 0.06]
Explain in detail about sampling error.
Explain the procedures of testing of hypothesis
State the different methods of measuring trend.
Construct the cost of living Index number for 2015 on the basis of 2012 from the following data using family budget method.
Commodity | Price | Weight | |
2012 | 2015 | ||
Rice | 250 | 280 | 10 |
Wheat | 70 | 280 | 5 |
Corn | 150 | 170 | 6 |
Oil | 25 | 35 | 4 |
Dhal | 85 | 90 | 3 |
Determine an initial basic feasible solution to the following transportation problem using North West corner rule.
Here Oi and Dj represent ith origin and jth destination.
Solve the following assignment problem.
If \(\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 2 \\ -1 \\ 3 \end{matrix} \right] \) find x,y and z
\(\int x^{3} \sin \left(x^{4}\right) d x\)
The supply for a commodity is \(\mathrm{p}=x^{2}+4 x+5\), where x denotes supply. Find the producer's surplus when price is 10.
Solve: \(\left(x^{2}-y x^{2}\right) d y+\left(y^{2}+x^{2} y^{2}\right) d x=0\)
Using Lagrange's formula, find the value of y when x = 42 from the following table
x | 40 | 50 | 60 | 70 |
y | 31 | 73 | 124 | 159 |
If a random variable. X has the probability distribution
X | 0 | 1 | 2 | 3 | 4 | 5 |
P(X=x) | a | 2a | 3a | 4a | 5a | 6a |
then find F(4)
Given the p.d.f of a continuous random variable X as follows \(f(x)= \begin{cases}k x(1-x) & \text { for } 0<x<1 \\ 0 & \text { otherwise }\end{cases}\). Find k and c.d.f.
The average percentage of failure in a certain examination is 40. What is the probàbility that out of a group of 6 candidates atleast 4 passed in the examination?
A random sample of 500 apples was taken from large consignment and 45 of them were found to be bad. Find the limits at which the bad apples lie at 99% confidence interval.
Compute Fisher's price index number for the following data.
Commodity | Base Year | Current Year | ||
Price | Quantity | Price | Quantity | |
A | 10 | 12 | 12 | 15 |
B | 7 | 15 | 5 | 20 |
C | 5 | 24 | 9 | 20 |
D | 16 | 5 | 14 | 5 |
Answers
Given
A = \(\left( \begin{matrix} 1 & 1 & -1 \\ 2 & -3 & 4 \\ 3 & -2 & 3 \end{matrix} \right) \) and B = \(\left( \begin{matrix} 1 & -2 & 3 \\ -2 & 4 & -6 \\ 5 & 1 & -1 \end{matrix} \right) \)
\(AB=\left( \begin{matrix} 1 & 1 & -1 \\ 2 & -3 & 4 \\ 3 & -2 & 3 \end{matrix} \right) \left( \begin{matrix} 1 & -2 & 3 \\ -2 & 4 & -6 \\ 5 & 1 & -1 \end{matrix} \right) \)
= \(\left( \begin{matrix} 1-2+5 & -2+4-1 & 3-6+1 \\ 2+6+20 & -4-12+45 & 6+18-4 \\ 3+4+15 & -6-8+3 & 9+12-3 \end{matrix} \right) \)
= \(\left( \begin{matrix} -6 & 1 & -2 \\ 28 & -12 & 20 \\ 22 & -11 & 18 \end{matrix} \right) =\left( \begin{matrix} -6 & 1 & -2 \\ 28 & -12 & 20 \\ 22 & -11 & 18 \end{matrix} \right) \)
Matrix (AB) | Elementary Transformation |
---|---|
\(AB=\left( \begin{matrix} -6 & 1 & -2 \\ 28 & -12 & 20 \\ 22 & -11 & 18 \end{matrix} \right) \) | |
\(\sim \left( \begin{matrix} 1 & -6 & -2 \\ -12 & 28 & 20 \\ -11 & 22 & 18 \end{matrix} \right) \) | \({ C }_{ 1 }\leftrightarrow { C }_{ 2 }\) |
\(\sim \left( \begin{matrix} 1 & -6 & -2 \\ -12 & 28 & 20 \\ -11 & 22 & 18 \end{matrix} \right) \) | \({ R }_{ 2 }\rightarrow { R }_{ 2 }+12{ R }_{ 1 }\) |
\(\sim \left( \begin{matrix} 1 & -6 & -2 \\ 0 & -44 & -4 \\ 0 & -44 & -4 \end{matrix} \right) \) | \({ R }_{ 3 }\rightarrow { R }_{ 3 }+11{ R }_{ 1 }\) |
\(\sim \left( \begin{matrix} 1 & -6 & -2 \\ 0 & -44 & -4 \\ 0 & 0 & 0 \end{matrix} \right) \) | \({ R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 2 }\) |
The matrix is in echelon form and the number of non-zero rows is 2.
\(\therefore \rho (AB)=2\)
Now \(BA=\left( \begin{matrix} 1 & -2 & 3 \\ -2 & 4 & -6 \\ 5 & 1 & -1 \end{matrix} \right) \left( \begin{matrix} 1 & 1 & -1 \\ 2 & -3 & 4 \\ 3 & -2 & 3 \end{matrix} \right) \)
= \(\left( \begin{matrix} 1-4+9 & 1+6-6 & -1-8+9 \\ -2+8-18 & -2-12+12 & 2+16-18 \\ 5+2-3 & 5-3+2 & -5+4-3 \end{matrix} \right) \)
= \(\left( \begin{matrix} 6 & 1 & 0 \\ -12 & -2 & 0 \\ 4 & 4 & -4 \end{matrix} \right) \)
Matrix (BA) | Elementary Transformation |
---|---|
\(BA=\left( \begin{matrix} 6 & 1 & 0 \\ -12 & -2 & 0 \\ 4 & 4 & -4 \end{matrix} \right) \) | |
\(\sim \left( \begin{matrix} 1 & 6 & 0 \\ -2 & -12 & 0 \\ 4 & 4 & -4 \end{matrix} \right) \) | \({ C }_{ 1 }\leftrightarrow { C }_{ 2 }\) |
\(\sim \left( \begin{matrix} 1 & 6 & 0 \\ 0 & 0 & 0 \\ 4 & 4 & -4 \end{matrix} \right) \) | \({ R }_{ 2 }\rightarrow { R }_{ 2 }+2{ R }_{ 1 }\) |
\(\sim \left( \begin{matrix} 1 & 6 & 0 \\ 0 & 0 & 0 \\ 0 & -20 & -4 \end{matrix} \right) \) | \({ R }_{ 3 }\rightarrow { R }_{ 3 }-4R_{ 1 }\) |
The number of non-zero rows is 2.
\(\therefore \rho (BA)=2\)
Let the hourly charge for horse riding be Rs. x and the hourly charge for quad bike be Rs. y from the given data,
3x + 4y = 780
2x+ 3y = 560
\(\Delta =\left| \begin{matrix} 3 & 4 \\ 2 & 3 \end{matrix} \right| =3(3)-2(4)=9-8=1\neq 0\)
Since \(\Delta \neq 0\) the system is consistent with unique solution and Cramer's rule can be applied
\(\Delta x=\left| \begin{matrix} 780 & 4 \\ 560 & 3 \end{matrix} \right| =780(30-4(560)\)
= 2340 - 2240 = 100
\(\Delta y=\left| \begin{matrix} 3 & 780 \\ 2 & 560 \end{matrix} \right| =39560)-2(780)\)
= \(1680 - 1560\) = 120
\(\therefore x=\cfrac { \Delta x }{ \Delta } =\cfrac { 100 }{ 1 } =100\)
\(y=\cfrac { \Delta y }{ \Delta } =\cfrac { 120 }{ 1 } =120\)
\(\therefore\) Hourly charges for the two rides are Rs. 100 and Rs. 120 respectively.
\(\int { \frac { 7x-1 }{ { x }^{ 2 }-5x+6 } dx } =\int { \left[ \frac { 20 }{ x-3 } -\frac { 13 }{ x-2 } \right] dx } \)
\(=20\int { \frac { dx }{ x-3 } -13\int { \frac { dx }{ x-2 } } } \)
\(20log\left| x-3 \right| -13 \log\left| x-2 \right| +c\)
[ By partial fractions,
\(\frac { 7x-1 }{ { x }^{ 2 }-5x+6 } =\frac { A }{ x-3 } +\frac { B }{ x-2 } \Rightarrow \frac { 7x-1 }{ { x }^{ 2 }-5x+6 } =\frac { 20 }{ x-3 } -\frac { 13 }{ x-2 } \)]
\(\int { \left( { x }^{ 2 }-2x+5 \right) } { e }^{ -x }dx =\int { udv } \)
= uv − u'v1 + u''v2 − u'''v3 + ...
= (x2 − 2x + 5)(−e−x )− (2x − 2)e−x + 2(−e−x )+ c
= e−x (−x2 −5)+ c
Successive derivatives | Repeated integrals |
Take u = x2 − 2x + 5 |
and dv = e−xdx v = − e−x v1 = e-x v2 = e-x |
Given cost of overhaul of an engine is Rs. 10,000
Operating cost per hour = 2x - 240.
Total cost for the engine to run for 300 hours
after overhaul =10,000+\(\int _{ 0 }^{ 300 }{ (2x-240) } dx\)
\(=10,000+{ \left[ \frac { { 2x }^{ 2 } }{ 2 } -240x \right] }_{ 0 }^{ 300 }\)
\(=10,000+{ [{ x }^{ 2 }-240x] }_{ 0 }^{ 300 }\)
= 10,000 + 90,000 - 72,000
= 1,00,000 - 72,000
= Rs. 28,000
Given MR = 6 − 3x2 − x3
⇒ ഽMR =ഽ(6 − 3x2 − x3)dx
\(\Rightarrow \mathrm{R}=6 x-\frac{\not{3} x^{3}}{\not3}-\frac{x^{4}}{4}+k\)
\(\Rightarrow 6x-{ x }^{ 3 }-\frac { { x }^{ 4 } }{ 4 } +k\)
When x = 0, R = 0 ⇒ k = 0
\(\Rightarrow R=6x-{ x }^{ 3 }-\frac { { x }^{ 4 } }{ 4 } \)
Demand function \(P=\frac { R }{ x } =6-{ x }^{ 2 }-\frac { { x }^{ 3 } }{ 4 } \)
Given equation can be written as \(\frac { ydx-xdy }{ { y }^{ 2 } } -{ 3x }^{ 2 }{ e }^{ { x }^{ 2 } }dx=0\)
Integrating, ഽ\(\frac { ydx-xdy }{ { y }^{ 2 } } \) - ഽ3x2ex3 dx = c
ഽ\(d\left( \frac { x }{ y } \right) \) - ഽetdt = c
(where t = x3 and dt = 3x2dx )
\(\frac { x }{ y } \) - et = c
\(\frac { x }{ y } \) - \({ e }^{ { x }^{ 2 } }\) = c
Given \(\frac { a+{ x }^{ 2 } }{ 1+y } =xy\frac { dy }{ dx } \)
Separating the variables we get,
\(\frac { (1+{ x }^{ 2 })dx }{ x } \) = y(1+y)dy
⇒ \(\left( \frac { 1 }{ x } +\frac { { x }^{ 2 } }{ x } \right) \)dx = (y+y2)dy
⇒ \(\left( \frac { 1 }{ x } +x \right) \) = (y+y2)dy
Integrating both sides we get,
\(\int { \left( \frac { 1 }{ x } +x \right) dx } =\int { (y+y^{ 2 })dy } \)
⇒ logx + \(\frac { { x }^{ 2 } }{ 2 } =\frac { { y }^{ 2 } }{ 2 } +\frac { y^{ 3 } }{ 3 } \) + C.
Since only four values of f(x) are given, the polynomial which fits the data is of degree three. Hence fourth differences are zeros.
(ie) Δ4y0 = 0, ∴ (E−1)4y = 0
(E4 - 4E3 + 6E2 - 4E +1)y0 = 0
E4y0 - 4E3y0+ 6E2y0 - 4Ey0 +y0 = 0
y4 − 4y3 + 6y2 − 4y1 + y0 = 0
67.4 − 4y3+ 6(54.1) − 4(4.2) + 45 = 0
240.2 = 4y3 ∴ y3 = 60.05
Scale:
In x-axis 1 cm = 100 units
In y-axis 1 cm = 1000 units
Plot the points (200, 2500), (300, 2800) (400, 3100) (640; 3820), (540,3220) and (580, 3640).
At x = 350, draw a vertical line and from the intersecting point on the curve, draw a horizontal line.
From the graph, we find that when x = 350, y = 2900.
Hence, the expense at a level of 350 units is Rs. 2900
Sample space \((s)=\left\{ \begin{matrix} (1,1) & (1,2) & (1,3) \\ (2,1) & (2,2) & (2,3) \\ \begin{matrix} (3,1) \\ (4,1) \\ \begin{matrix} (5,1) \\ (6,1) \end{matrix} \end{matrix} & \begin{matrix} (3,2) \\ (4,2) \\ \begin{matrix} (5,2) \\ (6,2) \end{matrix} \end{matrix} & \begin{matrix} (3,3) \\ (4,3) \\ \begin{matrix} (5,3) \\ (6,3) \end{matrix} \end{matrix} \end{matrix}\begin{matrix} (1,4) & (1,5) & (1,6) \\ (2,4) & (2,5) & (2,6) \\ \begin{matrix} (3,4) \\ (4,4) \\ \begin{matrix} (5,4) \\ (6,4) \end{matrix} \end{matrix} & \begin{matrix} (3,5) \\ (4,5) \\ \begin{matrix} (5,5) \\ (6,5) \end{matrix} \end{matrix} & \begin{matrix} (3,6) \\ (4,6) \\ \begin{matrix} (5,6) \\ (6,6) \end{matrix} \end{matrix} \end{matrix} \right\} \)
Total outcomes : n(S) = 36
We know that
\(\int _{ -\infty }^{ \infty }{ f(x)dx=1 } \),since f(x) is a density function
\(\int _{ 0 }^{ \infty }{ { ke }^{ -2x } } dx=1\)
\(k\int _{ 0 }^{ \infty }{ { ke }^{ -2x } } dx=1\)
\(k{ \left[ \frac { { e }^{ -2x } }{ -2 } \right] }_{ 0 }^{ \infty }=1\)
⇒ k = 2
\(E(X)=\int _{ -\infty }^{ \infty }{ xf(x)dx } \)
\(=\int _{ 0 }^{ \infty }{ x{ e }^{ -2x } } dx\)
\(=2\int _{ 0 }^{ \infty }{ { xe }^{ -2x }dx } \)
\(=2\left\{ { \left[ \frac { { xe }^{ -2x } }{ -2 } \right] }_{ 0 }^{ \infty }\int _{ 0 }^{ \infty }{ \frac { { e }^{ -2x } }{ -2 } dx } \right\} \)
\((\because \int { udv=uv-\int { udv } } )\)
\(=\int _{ 0 }^{ \infty }{ { e }^{ -2x }dx } \) \(=\frac { 1 }{ 2 } \)
Let X be a random variable follows poisson distribution with number of erythrocytes.
Hence, Mean λ = 8 erythrocytes/single square
P(exactly 5 erythrocytes are in one square) = P(X = 5)=\(\\ \frac { { e }^{ -\lambda }{ \lambda }^{ x } }{ x! } =\frac { { e }^{ -8 }{ 8 }^{ 5 } }{ 5! } \)
\(=\frac { 0.000335\times 32768 }{ 120 } \)
= 0.0916
The probability that exactly 5 erythrocytes are found in one square is 0.0916. i.e there are 9.16% chances that exactly 5 erythrocytes are found in one square.
Let X be probability of mortality rate for a certain disease
∴ p = \(\frac { 7 }{ 1000 } \) and n = 400
Hence X follows poisson distribution with p(x, λ)
= \(\frac { e^{ -\lambda }.{ \lambda }^{ x } }{ x! } \)
∴ P (just 2 deaths) = P(X = 2)
= \(\frac { e^{ -2.8 }(2.8)^{ 2 } }{ 2! } \) [∵ p(x,λ) =\(\frac { e^{ -\lambda }.{ \lambda }^{ x } }{ x! } \), λ = 2.8, x = 2]
= \(\frac { (0.06)(2.8)^{ 2 } }{ 2 } \) = (0.03)(2.8)2 = 0.2352
∴ Probability for just 2 deaths on account of this disease = 0.2352.
Errors, which arise in the normal course of investigation or enumeration on account of chance, are called sampling errors. Sampling Errors arise due to the following reasons:
a) Faulty selection of the sample instead of correct sample by defective sampling technique.
b) The investigator substitutes a convenient sample if the original sample is not available while investigation.
c) In area surveys, while dealing with border lines it depends upon the investigator whether to include them in the sample or not. This is known as Faulty demarcation of sampling units.
Hypothesis testing is also referred as "Statistical Decision Making".
There are two types of statistical hypothesis
(i) Null hypothesis: which is tested for possible rejection under the assumption that it is true
(ii) Alternative hypothesis which is complementary to the null hypothesis
For example: Ho:μ=μ0
i) H1:μ≠μ0 is know as two tailed alternative test
ii) H1:μ>μ0 and H1:μ<μ0 are known as one tailed alternative
iii) H1:μ>μ0 is said to be right tailed test where the critical region lies entirely on the right tail of the normal curve.
iv) H1:μ<μ0 is said to the left tailed test where the critical region lies entirely on the left tail of the normal curve.
The different methods of measurements of trends are
(i) Free hand or graphic method
(ii) Method of semi averages
(iii) Method of moving averages
(iv) Method of least squares
Commodity | Price | P = \(\frac {p_{1}}{p_{0}}\) \(\times 100\) | V | [PV | |
2012 (p0) | 2015 (p0) | ||||
Rice | 250 | 280 | 112 | 10 | 1120 |
Wheat | 70 | 85 | 121.42 | 5 | 607.1 |
Corn | 150 | 170 | 113.33 | 6 | 679.98 |
Oil | 25 | 35 | 140 | 4 | 560 |
Dhal | 85 | 90 | 105.88 | 3 | 317.64 |
28 | 3284.72 |
Using family budget method,
C.L.I = \(\frac {\sum PV}{\sum V}\) = \(\frac {3284.72}{28}\) = 117.31
Given transportation table is
Total Availability = Total Requirement
Therefore the given problem is balanced transportation problem.
Hence there exists a feasible solution to the given problem.
First allocation :
Second allocation :
Third Allocation :
Fourth Allocation :
Fifth allocation :
Final allocation :
Transportation schedule : O1⟶D1, O1⟶D2, O2⟶D2, O2⟶D3, O3⟶D3,O3⟶D3.
The transportation cost
= (6\(\times\)6)+(8\(\times\)4)+(2\(\times\)9)+(14\(\times\)2)+(1\(\times\)6)+(4\(\times\)2)
= Rs.128
Since the number of columns is less than the number of rows, given assignment problem is unbalanced one. To balance it , introduce a dummy column with all the entries zero. The revised assignment problem is
Here only 3 tasks can be assigned to 3 men.
Step 1: Its not necessary, since each row contains zero entry. Go to Step 2.
Step 2:
Step 3 (Assignment) :
Since each row and each columncontains exactly one assignment,all the three men have been assigned a task. But task S is not assigned to any Man. The optimal assignment schedule and total cost is
Task | Men | cost |
P | 1 | 9 |
Q | 3 | 6 |
R | 2 | 20 |
s | d | 0 |
Total cost | 35 |
The optimal assignment (minimum) cost = Rs. 35
\(\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 2 \\ -1 \\ 3 \end{matrix} \right] \)
\(\Rightarrow \left( \begin{matrix} x0+0 \\ 0+0+z \\ 0+y+0 \end{matrix} \right) =\left( \begin{matrix} 2 \\ -1 \\ 3 \end{matrix} \right) \)
\(\Rightarrow \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) =\left( \begin{matrix} 2 \\ -1 \\ 3 \end{matrix} \right) \)
\(\Rightarrow x=2\quad z=-1\quad y=3\)
\(\therefore\) Solution set is {2,3, -1}
\(
\mathrm{t} =x^{4}, \mathrm{dt}=4 x^{3} \mathrm{~d} x
\)
\(\frac{d t}{4} =x^{3} \mathrm{~d} x
\)
\(\frac{1}{4} \int \sin t d t =\frac{-1}{4} \cos t+c
\)
\( =-\frac{1}{4} \cos \left(\mathrm{x}^{4}\right)+c
\)
\(
\mathrm{p} =x^{2}+4 x+5
\)
\(\mathrm{p} =10
\)
\(x^{2}+4 x+5 =10
\)
\(x^{2}+4 x-5 =0
\)
\((x+5)(x-1) =0 \Rightarrow x=1
\)
\(
\mathrm{PS} =p_{0} x_{0}-\int_{0}^{x_{1}} g(x) d x
\)
\( =36-\int_{0}^{3}\left(3+x^{2}\right) d x=36-\left[3 x+\frac{x^{3}}{3}\right]_{0}^{3}
\)
\(=36-[9+9]=36-18=18 \text { units }\)
\(
\left(x^{2}-\mathrm{y} x^{2}\right) \mathrm{dy} =-\left(\mathrm{y}^{2}+x^{2} \mathrm{y}^{2}\right) \mathrm{d} x
\)
\(x^{2}(1-\mathrm{y}) \mathrm{dy} =-\mathrm{y}^{2}\left(1+x^{2}\right) \mathrm{d} x
\)
\(\int \frac{1-y}{y^{2}} d y =-\int \frac{\left(1+x^{2}\right)}{x^{2}} d x
\)
\(\int \frac{d y}{y^{2}}-\int \frac{y d y}{y^{2}} =-\int x^{-2} d x-\int d x
\)
\(\int y^{-2} d y-\int \frac{d y}{y} =-\int x^{-2} d x-\int d x
\)
\(\frac{y^{-1}}{-1}-\log |y| =\frac{x^{-1}}{1}-x+c
\)
\(\log |y|+\frac{1}{y}+\frac{1}{x}-x =\mathrm{c}
\)
By data, we have
xo = 40, x1 = 50, x2 = 60, x3 = 70
yo = 31, y1 = 73, y2 = 124, y3 = 159.
Using Lagrange's formula, we get
\(y={ y }_{ 0 }\frac { (x-{ x }_{ 1 })(x-{ x }_{ 2 })(x-{ x }_{ 3 }) }{ ({ x }_{ 0 }-{ x }_{ 1 })({ x }_{ 0 }-{ x }_{ 2 })({ x }_{ 0 }-{ x }_{ 3 }) } +{ y }_{ 1 }\frac { (x-{ x }_{ 0 })(x-{ x }_{ 2 })(x-{ x }_{ 3 }) }{ ({ x }_{ 1 }-{ x }_{ 0 })({ x }_{ 1 }-{ x }_{ 2 })({ x }_{ 1 }-{ x }_{ 3 }) } { y }_{ 2 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 3 }) }{ ({ x }_{ 2 }-{ x }_{ 0 })({ x }_{ 2 }-{ x }_{ 1 })({ x }_{ 2 }-{ x }_{ 3 }) } +{ y }_{ 3 }\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 3 }) }{ ({ x }_{ 3 }-{ x }_{ 0 })({ x }_{ 3 }-{ x }_{ 1 })({ x }_{ 3 }-{ x }_{ 2 }) } \)
∴ y(42) = 31\(\frac { (-8)(-18)(-28) }{ (-10)(-20)(-30) } +73\frac { (2)(-18)(-28) }{ (10)(-10)(-20) } +124\frac { (2)(-8)(-28) }{ (20)(10)(-10) } +59\frac { (2)(-8)(-28) }{ (30)(20)(10) } \)
= 20. 832 + 36. 792 - 27. 776 + 7.632
y = 37. 48
Since the random variable X is the probability distribution function, Σpi = 1
∴ a + 2a + 3a + 4a + 5a + 6a = 1
21a = 1 ⇒ a = \(\frac{1}{21}\)
Now, F(4) = P(X ≤ 4)
= P(X = 0) + P(X = 1) + P(X = 2)P(X = 3) + P(X = 4)
= a + 2a + 3a + 4a + 5a = 15a
= 15\((\frac{1}{21})=\frac{5}{7}\)
∴ F(4) = \(\frac{5}{7}\)
\(\because\) if is a p.d.f
\(
\int_{-\infty}^{x} f(x) d x =1
\)
\(k \int_{0}^{1} x(1-x) d x =1
\)
\(k \int_{0}^{1}\left(x-x^{2}\right) d x =1
\)
\(k\left[\frac{x^{2}}{2}-\frac{x^{3}}{3}\right]_{0}^{1} =1
\)
\(k\left(\frac{1}{2}-\frac{1}{3}\right) =1
\)
\(k\left(\frac{1}{6}\right)=1 \Rightarrow k=6\)
Cumulative distribution function
\(
\mathrm{F}(x) =0 \text { for } x \geq 0
\)
\(\mathrm{F}(x) =\mathrm{P}(\mathrm{X} \leq x)=\int_{-1}^{1} f(x) d x
\)
\( =\int_{0}^{x} 6 x(1-x) d x
\)
\(\int_{0}^{x}\left(6 x-6 x^{2}\right) d x =\left[\frac{6 x^{2}}{2}-\frac{6 x^{3}}{3}\right]_{0}^{x}=3 x^{2}-2 x^{3}
\)
\(\therefore \mathrm{F}(x) =0 \text { for } x<0
\)
\( =3 x^{2}-2 x^{3} \text { for } 0<x<1 \)
\( =1 \text { for } x \geq 1
\)
Let X be the random variable denoting the number of student who passes.
\(
\mathrm{p}=\frac{60}{100}=\frac{3}{5}, \mathrm{q}=\frac{40}{100}=\frac{2}{5}
\)
\(\mathrm{n}=6, \mathrm{P}(\mathrm{X}=\mathrm{x})=n C_{x} p^{x} q^{n-x}
\)
\(\mathrm{P}^{\prime}(\mathrm{X} \geq 4)=\mathrm{P}(\mathrm{X}=4)+\mathrm{P}(\mathrm{X}=5)+\mathrm{P}(\mathrm{X}=6)\)
\(
=6 C_{4}\left(\frac{3}{5}\right)^{4}\left(\frac{2}{5}\right)^{2}+6 C_{1}\left(\frac{3}{5}\right)^{4}\left(\frac{2}{5}\right)+\left(\frac{3}{5}\right)^{6} \)
\(=15 \frac{(81 \times 4)}{5^{6}}+\frac{(6)(243)(2)}{5^{6}}+\frac{729}{5^{6}}
\)
\( =\frac{8505}{15625}\)
n = 500
P = proportion of bad apples
\(=\frac{45}{500}=0.09\)
q = 1 - p = 1 - 0.09 = 0.91
Confidence limits for the population p of bad apples are given by
\(=p \pm Z_{c} \sqrt{\frac{p q}{n}}\)
\(=0.09 \pm(2.58) \sqrt{\frac{(0.09)(0.91)}{500}}\)
\(=0.09 \pm 0.033\)
Required interval is (0.057, 0.123) Thus the bad apples in the consignment lie between 5.7% and 12.3%
Commodity | Base Year | Current Year | ||
p0 | q0 | p1 | q1 | |
A | 10 | 12 | 12 | 15 |
B | 7 | 15 | 5 | 20 |
C | 5 | 24 | 9 | 20 |
D | 16 | 5 | 14 | 5 |
p1q0 | p0q0 | p1q1 | p0q1 |
144 | 120 | 180 | 150 |
75 | 105 | 100 | 140 |
216 | 120 | 180 | 100 |
70 | 80 | 70 | 80 |
505 | 425 | 530 | 470 |
Fisher's ideal index = \(\sqrt \frac {\sum p_{1}q_{0}\times \sum p_{1}q_{1}}{\sum p_{0}q_{0}\times \sum p_{0}q_{1}} \times 100\)
= \(\sqrt{\frac{505}{425} \times \frac {430}{470}} \times 100\)
\(P^{F}_{01}\) = 115.75