110 marksNumericalMeasures of dispersionAnswer

What are different methods of measuring dispersion?

Sample of polythene bags from two manufacturers, A, B, are tested by a prospective buyer for bursting pressure and the results are as follows. Which set of bags has more uniform pressure? If prices are the same, which manufacturer's bags would be preferred by buyer? Use appropriate statistical tool.

Bursting Pressure	5-10	10-15	15-20	20-25	25-30	30-35
A	2	9	29	54	11	5
B	9	11	18	32	27	13

[10]

Methods of Measuring Dispersion and Comparison of Bursting Pressure

Given Data

Bursting Pressure	Mid-value $x$	$f_A$	$f_B$
5-10	7.5	2	9
10-15	12.5	9	11
15-20	17.5	29	18
20-25	22.5	54	32
25-30	27.5	11	27
30-35	32.5	5	13

$\sum f_A = 110$ , $\sum f_B = 110$

Part 1: Methods of Measuring Dispersion

Absolute Measures

Range: $R = L - S$
Quartile Deviation: $QD = \dfrac{Q_3 - Q_1}{2}$
Mean Deviation: average of absolute deviations from mean/median
Standard Deviation & Variance: $\sigma = \sqrt{\dfrac{\sum f(x-\bar{x})^2}{N}}$ , $\sigma^2$

Relative Measures

Coefficient of Range $= \dfrac{L-S}{L+S}$
Coefficient of QD $= \dfrac{Q_3-Q_1}{Q_3+Q_1}$
Coefficient of MD $= \dfrac{MD}{\bar{x}}$
Coefficient of Variation: $CV = \dfrac{\sigma}{\bar{x}}\times 100$

To compare uniformity/consistency, the Coefficient of Variation (CV) is the appropriate tool. Lower CV means more uniform (consistent).

Part 2: Computation

Let assumed mean $A = 22.5$ , $h = 5$ , $d = \dfrac{x-22.5}{5}$

$x$	$d$	$f_A$	$f_A d$	$f_A d^2$	$f_B$	$f_B d$	$f_B d^2$
7.5	-3	2	-6	18	9	-27	81
12.5	-2	9	-18	36	11	-22	44
17.5	-1	29	-29	29	18	-18	18
22.5	0	54	0	0	32	0	0
27.5	1	11	11	11	27	27	27
32.5	2	5	10	20	13	26	52
Total		110	-32	114	110	-14	222

Manufacturer A

$\bar{x}_A = 22.5 + 5\times\frac{-32}{110} = 22.5 - 1.4545 = 21.045$

$\sigma_A = 5\sqrt{\frac{114}{110} - \left(\frac{-32}{110}\right)^2} = 5\sqrt{1.03636 - 0.08471}$

$\sigma_A = 5\sqrt{0.95165} = 5\times 0.97552 = 4.878$

$CV_A = \frac{4.878}{21.045}\times 100 = 23.18\%$

Manufacturer B

$\bar{x}_B = 22.5 + 5\times\frac{-14}{110} = 22.5 - 0.6364 = 21.864$

$\sigma_B = 5\sqrt{\frac{222}{110} - \left(\frac{-14}{110}\right)^2} = 5\sqrt{2.01818 - 0.01620}$

$\sigma_B = 5\sqrt{2.00198} = 5\times 1.41492 = 7.075$

$CV_B = \frac{7.075}{21.864}\times 100 = 32.36\%$

Conclusion

Manufacturer	Mean	SD	CV
A	21.05	4.878	23.18%
B	21.86	7.075	32.36%

Since $CV_A < CV_B$ , Manufacturer A's bags have more uniform (consistent) bursting pressure.

As prices are the same, the buyer should prefer Manufacturer A's bags because they show greater uniformity/consistency in bursting pressure.

Methods of Measuring Dispersion and Comparison of Bursting Pressure

Given Data

Bursting Pressure	Mid-value $x$	$f_A$	$f_B$
5-10	7.5	2	9
10-15	12.5	9	11
15-20	17.5	29	18
20-25	22.5	54	32
25-30	27.5	11	27
30-35	32.5	5	13

$\sum f_A = 110$, $\sum f_B = 110$

Part 1: Methods of Measuring Dispersion

Absolute Measures

Range: $R = L - S$
Quartile Deviation: $QD = \dfrac{Q_3 - Q_1}{2}$
Mean Deviation: average of absolute deviations from mean/median
Standard Deviation & Variance: $\sigma = \sqrt{\dfrac{\sum f(x-\bar{x})^2}{N}}$, $\sigma^2$

Relative Measures

Coefficient of Range $= \dfrac{L-S}{L+S}$
Coefficient of QD $= \dfrac{Q_3-Q_1}{Q_3+Q_1}$
Coefficient of MD $= \dfrac{MD}{\bar{x}}$
Coefficient of Variation: $CV = \dfrac{\sigma}{\bar{x}}\times 100$

To compare uniformity/consistency, the Coefficient of Variation (CV) is the appropriate tool. Lower CV means more uniform (consistent).

Part 2: Computation

Let assumed mean $A = 22.5$, $h = 5$, $d = \dfrac{x-22.5}{5}$

$x$	$d$	$f_A$	$f_A d$	$f_A d^2$	$f_B$	$f_B d$	$f_B d^2$
7.5	-3	2	-6	18	9	-27	81
12.5	-2	9	-18	36	11	-22	44
17.5	-1	29	-29	29	18	-18	18
22.5	0	54	0	0	32	0	0
27.5	1	11	11	11	27	27	27
32.5	2	5	10	20	13	26	52
Total		110	-32	114	110	-14	222

Manufacturer A

$$\bar{x}_A = 22.5 + 5\times\frac{-32}{110} = 22.5 - 1.4545 = 21.045$$

$$\sigma_A = 5\sqrt{\frac{114}{110} - \left(\frac{-32}{110}\right)^2} = 5\sqrt{1.03636 - 0.08471}$$

$$\sigma_A = 5\sqrt{0.95165} = 5\times 0.97552 = 4.878$$

$$CV_A = \frac{4.878}{21.045}\times 100 = 23.18%$$

Manufacturer B

$$\bar{x}_B = 22.5 + 5\times\frac{-14}{110} = 22.5 - 0.6364 = 21.864$$

$$\sigma_B = 5\sqrt{\frac{222}{110} - \left(\frac{-14}{110}\right)^2} = 5\sqrt{2.01818 - 0.01620}$$

$$\sigma_B = 5\sqrt{2.00198} = 5\times 1.41492 = 7.075$$

$$CV_B = \frac{7.075}{21.864}\times 100 = 32.36%$$

Conclusion

Manufacturer	Mean	SD	CV
A	21.05	4.878	23.18%
B	21.86	7.075	32.36%

Since $CV_A < CV_B$, Manufacturer A's bags have more uniform (consistent) bursting pressure.

As prices are the same, the buyer should prefer Manufacturer A's bags because they show greater uniformity/consistency in bursting pressure.

210 marksNumericalRegression AnalysisAnswer

Write the properties of correlation coefficient. The time it takes to transmit a file always depends on the file size. Suppose you transmitted 30 files, with the average size of 126 Kbytes and the standard deviation of 35 Kbytes. The average transmitted time was 0.04 seconds with the standard deviation 0.01 seconds. The correlation coefficient between the time and size was 0.86. Based on these data, fit a linear regression model and predict the time it take to transmit a 400Kbyte file.[10]

Let $X$ = file size (Kbytes), $Y$ = transmission time (seconds). Parameter X (size) Y (time) ------------------------------- Number of files $n = 30$ Mean $\bar{X} = 126$ $\bar{Y} = 0.04$ Std. deviation $SX = 35$ $SY = 0.01$ Correlation ...

310 marksDiscrete distributionsAnswer

a. What do you understand by Poisson distribution? What are its main features? b. What do you mean by joint probability distribution function? Write down its properties.[10]

--- Poisson distribution is a discrete probability distribution that is used to model the number of times an event occurs in a fixed interval of time or space, under the following conditions: - The probability of success is very small, i...

45 marksNumericalMeasures of central tendencyAnswer

If 50 image of your website, 10 have black and white image, and their average scanned image occupies with 2.5 megabytes of memory. The total image occupies by the entire work 281 megabytes. Find the average occupies megabytes of those color images. [5]

Parameter Value ------------------ Total images 50 Black and white (B&W) images 10 Average size of each B&W image 2.5 MB Total memory occupied by all images 281 MB --- $\text{Color images} = 50 - 10 = 40$ $$= 10 \times 2.5 = 25 \text{ ...

Parameter Value ------------------ Total images 50 Black and white (B&W) images 10 Average size of each B&W image 2.5 MB Total memory occupied by all images 281 MB --- $$\text{Color images} = 50 - 10 = 40$$ $$= 10 \times 2.5 = 25 \text{ ...

55 marksNumericalMeasures of central tendencyAnswer

Calculate Q1, D7 and P58 from the following data and interpret the results.

Weight	0-10	10-15	20-25	25-30	30-35	35-40	40-45	45-50	50-60
No. of person	4	8	30	15	13	6	4	4	1

[5]

Weight 0-10 10-15 20-25 25-30 30-35 35-40 40-45 45-50 50-60 ---------------------------------------------------------------------- No. of persons 4 8 30 15 13 6 4 4 1 Note: The class 15-20 is absent from the data. To keep the classes con...

65 marksNumericalJoint probability distribution of two randAnswer

Fatigue Test Probability Analysis

The following joint probability data apply to a fatigue test run on bronze strips. X represents cycles to failure (in $10^5$ ) when alternate strips are bent at a high level of deflection. Y represents the same at a lower deflection event.

$\begin{array}{c|cccc} X/Y & 20 & 30 & 40 & 50 \\ \hline 4 & 0.01 & 0.03 & 0.05 & 0.02 \\ 5 & 0.03 & 0.10 & 0.08 & 0.04 \\ 6 & 0.02 & 0.08 & 0.12 & 0.11 \\ 7 & 0.02 & 0.04 & 0.07 & 0.18 \\ \end{array}$

a. Find the marginal probability distribution for X and Y

b. Determine the conditional probability distribution of Y given $X = 5$

c. Are X and Y independent?

[5]

Fatigue Test Probability Analysis

The following joint probability data apply to a fatigue test run on bronze strips. X represents cycles to failure (in $10^5$) when alternate strips are bent at a high level of deflection. Y represents the same at a lower deflection event.

$$\begin{array}{c|cccc} X/Y & 20 & 30 & 40 & 50 \ \hline 4 & 0.01 & 0.03 & 0.05 & 0.02 \ 5 & 0.03 & 0.10 & 0.08 & 0.04 \ 6 & 0.02 & 0.08 & 0.12 & 0.11 \ 7 & 0.02 & 0.04 & 0.07 & 0.18 \ \end{array}$$

a. Find the marginal probability distribution for X and Y

b. Determine the conditional probability distribution of Y given $X = 5$

c. Are X and Y independent?

[5]

Joint probability table $f(x,y)$ : X/Y 20 30 40 50 ------------------------- 4 0.01 0.03 0.05 0.02 5 0.03 0.10 0.08 0.04 6 0.02 0.08 0.12 0.11 7 0.02 0.04 0.07 0.18 Marginal of X: $P(X=x)=\sumy f(x,y)$ $P(X=4)=0.01+0.03+0.05+0.02=0.11$ ...

Joint probability table $f(x,y)$: X/Y 20 30 40 50 ------------------------- 4 0.01 0.03 0.05 0.02 5 0.03 0.10 0.08 0.04 6 0.02 0.08 0.12 0.11 7 0.02 0.04 0.07 0.18 Marginal of X: $P(X=x)=\sumy f(x,y)$ $$P(X=4)=0.01+0.03+0.05+0.02=0.11$$ ...

75 marksNumericalDiscrete distributionsAnswer

Fit a binomial distribution to the following data:

$\begin{array}{c|ccccccc} X & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline f & 5 & 8 & 15 & 14 & 10 & 6 & 2 \\ \end{array}$

[5]

Fit a binomial distribution to the following data:

$$\begin{array}{c|ccccccc} X & 0 & 1 & 2 & 3 & 4 & 5 & 6 \ \hline f & 5 & 8 & 15 & 14 & 10 & 6 & 2 \ \end{array}$$

[5]

X 0 1 2 3 4 5 6 ------------------------ f 5 8 15 14 10 6 2 Maximum value of X is 6, so $n = 6$ . X f fX ---------- 0 5 0 1 8 8 2 15 30 3 14 42 4 10 40 5 6 30 6 2 12 Total N = 60 ΣfX = 162 $$\bar{X} = \frac{\Sigma fX}{N} = \frac{162}{60} ...

X 0 1 2 3 4 5 6 ------------------------ f 5 8 15 14 10 6 2 Maximum value of X is 6, so $n = 6$. X f fX ---------- 0 5 0 1 8 8 2 15 30 3 14 42 4 10 40 5 6 30 6 2 12 Total N = 60 ΣfX = 162 $$\bar{X} = \frac{\Sigma fX}{N} = \frac{162}{60} ...

85 marksNumericalJoint probability distribution of two randAnswer

If two random variables have the joint probability density function find (i) constant k (ii) conditional probability density function of X (iii) Identify whether X and Y are independent.

$f(x,y) = \begin{cases} k(2x + 3y), & 0 \leq x \leq 1, 0 \leq y \leq 1 \\ 0, & \text{otherwise} \end{cases}$

[5]

If two random variables have the joint probability density function find (i) constant k (ii) conditional probability density function of X (iii) Identify whether X and Y are independent.

$$f(x,y) = \begin{cases} k(2x + 3y), & 0 \leq x \leq 1, 0 \leq y \leq 1 \ 0, & \text{otherwise} \end{cases}$$

[5]

Joint PDF: $f(x,y) = k(2x + 3y)$ , $0 \le x \le 1$ , $0 \le y \le 1$

Given data

$f(x,y) = k(2x+3y)$ for $0 \le x \le 1$ , $0 \le y \le 1$ ; zero otherwise.

(i) Constant k

Normalization condition:

$\int_0^1 \int_0^1 k(2x+3y)\,dx\,dy = 1$

Inner integral over $x$ :

$\int_0^1 (2x+3y)\,dx = \left[x^2 + 3xy\right]_0^1 = 1 + 3y$

Outer integral over $y$ :

$k\int_0^1 (1+3y)\,dy = k\left[y + \tfrac{3y^2}{2}\right]_0^1 = k\left(1+\tfrac{3}{2}\right) = \tfrac{5}{2}k$

Set equal to 1:

$\tfrac{5}{2}k = 1 \implies \boxed{k = \tfrac{2}{5}}$

(ii) Conditional PDF of X (given Y)

Marginal of $Y$ :

$f_Y(y) = \int_0^1 \tfrac{2}{5}(2x+3y)\,dx = \tfrac{2}{5}(1+3y), \quad 0 \le y \le 1$

Conditional PDF:

$f_{X|Y}(x|y) = \frac{f(x,y)}{f_Y(y)} = \frac{\tfrac{2}{5}(2x+3y)}{\tfrac{2}{5}(1+3y)}$

$\boxed{f_{X|Y}(x|y) = \frac{2x+3y}{1+3y}, \quad 0 \le x \le 1}$

(iii) Independence

Marginal of $X$ :

$f_X(x) = \int_0^1 \tfrac{2}{5}(2x+3y)\,dy = \tfrac{2}{5}\left[2xy + \tfrac{3y^2}{2}\right]_0^1 = \tfrac{2}{5}\left(2x + \tfrac{3}{2}\right) = \frac{4x+3}{5}$

Product of marginals:

$f_X(x)\,f_Y(y) = \frac{4x+3}{5}\cdot\frac{2(1+3y)}{5} = \frac{2(4x+3)(1+3y)}{25}$

Compare with joint:

$f(x,y) = \frac{2(2x+3y)}{5}$

Since

$\frac{2(4x+3)(1+3y)}{25} \neq \frac{2(2x+3y)}{5},$

the joint density does not factor into the product of marginals.

Conclusion: X and Y are NOT independent.

Joint PDF: $f(x,y) = k(2x + 3y)$, $0 \le x \le 1$, $0 \le y \le 1$

Given data

$f(x,y) = k(2x+3y)$ for $0 \le x \le 1$, $0 \le y \le 1$; zero otherwise.

(i) Constant k

Normalization condition:

$$\int_0^1 \int_0^1 k(2x+3y),dx,dy = 1$$

Inner integral over $x$:

$$\int_0^1 (2x+3y),dx = \left[x^2 + 3xy\right]_0^1 = 1 + 3y$$

Outer integral over $y$:

$$k\int_0^1 (1+3y),dy = k\left[y + \tfrac{3y^2}{2}\right]_0^1 = k\left(1+\tfrac{3}{2}\right) = \tfrac{5}{2}k$$

Set equal to 1:

$$\tfrac{5}{2}k = 1 \implies \boxed{k = \tfrac{2}{5}}$$

(ii) Conditional PDF of X (given Y)

Marginal of $Y$:

$$f_Y(y) = \int_0^1 \tfrac{2}{5}(2x+3y),dx = \tfrac{2}{5}(1+3y), \quad 0 \le y \le 1$$

Conditional PDF:

$$f_{X|Y}(x|y) = \frac{f(x,y)}{f_Y(y)} = \frac{\tfrac{2}{5}(2x+3y)}{\tfrac{2}{5}(1+3y)}$$

$$\boxed{f_{X|Y}(x|y) = \frac{2x+3y}{1+3y}, \quad 0 \le x \le 1}$$

(iii) Independence

Marginal of $X$:

$$f_X(x) = \int_0^1 \tfrac{2}{5}(2x+3y),dy = \tfrac{2}{5}\left[2xy + \tfrac{3y^2}{2}\right]_0^1 = \tfrac{2}{5}\left(2x + \tfrac{3}{2}\right) = \frac{4x+3}{5}$$

Product of marginals:

$$f_X(x),f_Y(y) = \frac{4x+3}{5}\cdot\frac{2(1+3y)}{5} = \frac{2(4x+3)(1+3y)}{25}$$

Compare with joint:

$$f(x,y) = \frac{2(2x+3y)}{5}$$

Since

$$\frac{2(4x+3)(1+3y)}{25} \neq \frac{2(2x+3y)}{5},$$

the joint density does not factor into the product of marginals.

Conclusion: X and Y are NOT independent.

95 marksNumericalMomentsAnswer

Compute first four moments about arbitrary point 4 from following distribution and describe the characteristics of data.

$\begin{array}{c|ccccc} X & 2 & 3 & 4 & 5 & 6 \\ \hline f & 1 & 3 & 7 & 2 & 1 \\ \end{array}$

[5]

Compute first four moments about arbitrary point 4 from following distribution and describe the characteristics of data.

$$\begin{array}{c|ccccc} X & 2 & 3 & 4 & 5 & 6 \ \hline f & 1 & 3 & 7 & 2 & 1 \ \end{array}$$

[5]

X 2 3 4 5 6 ------------------ f 1 3 7 2 1 $N = \sum f = 14$ , Arbitrary point $A = 4$ . Moment formula about $A$ : $\mur' = \dfrac{\sum f d^r}{N}$ , where $d = X - A = X - 4$ . X f d fd fd² fd³ fd⁴ ---------------------------- 2 1 -2 -2 4 -8...

X 2 3 4 5 6 ------------------ f 1 3 7 2 1 $N = \sum f = 14$, Arbitrary point $A = 4$. Moment formula about $A$: $\mur' = \dfrac{\sum f d^r}{N}$, where $d = X - A = X - 4$. X f d fd fd² fd³ fd⁴ ---------------------------- 2 1 -2 -2 4 -8...

105 marksNumericalContinuous distributionAnswer

The lifetime of a certain electronic component is a normal random variate with the expectation of 5000 hours and a standard deviation of 100 hours. Compute the probabilities under the following conditions:a. Lifetime of components between 3000 to 6500 hours b. Lifetime of components between 3000 to 6500 hours c. Lifetime of components more than 6000 hours [5]

Mean $\mu = 5000$ hours - Standard deviation $\sigma = 100$ hours - $X \sim N(5000, 100^2)$ Standardization: $Z = \frac{X - \mu}{\sigma} = \frac{X - 5000}{100}$ Note on the question: Parts (a) and (b) are stated identically ("between...

Mean $\mu = 5000$ hours - Standard deviation $\sigma = 100$ hours - $X \sim N(5000, 100^2)$ Standardization: $$Z = \frac{X - \mu}{\sigma} = \frac{X - 5000}{100}$$ Note on the question: Parts (a) and (b) are stated identically ("between...

115 marksNumericalSpearman's rank correlationAnswer

Calculate Spearman's rank correlation coefficient for the following ranks given by three judges in a music contest. Indicate which pair of judges has the nearest approaches to music.

$\begin{array}{c|cccccccccc} \text{1}^{\text{st}}\text{ Judge} & 2 & 1 & 4 & 6 & 5 & 8 & 9 & 10 & 7 & 3 \\ \text{2}^{\text{nd}}\text{ Judge} & 4 & 3 & 2 & 5 & 1 & 6 & 8 & 9 & 10 & 7 \\ \text{3}^{\text{rd}}\text{ Judge} & 5 & 8 & 4 & 7 & 10 & 2 & 1 & 6 & 9 & 3 \\ \end{array}$

[5]

Calculate Spearman's rank correlation coefficient for the following ranks given by three judges in a music contest. Indicate which pair of judges has the nearest approaches to music.

$$\begin{array}{c|cccccccccc} \text{1}^{\text{st}}\text{ Judge} & 2 & 1 & 4 & 6 & 5 & 8 & 9 & 10 & 7 & 3 \ \text{2}^{\text{nd}}\text{ Judge} & 4 & 3 & 2 & 5 & 1 & 6 & 8 & 9 & 10 & 7 \ \text{3}^{\text{rd}}\text{ Judge} & 5 & 8 & 4 & 7 & 10 & 2 & 1 & 6 & 9 & 3 \ \end{array}$$

[5]

$n = 10$ competitors. Competitor 1st Judge ( $R1$ ) 2nd Judge ( $R2$ ) 3rd Judge ( $R3$ ) :---::---::---::---: 1 2 4 5 2 1 3 8 3 4 2 4 4 6 5 7 5 5 1 10 6 8 6 2 7 9 8 1 8 10 9 6 9 7 10 9 10 3 7 3 Formula: $$rs = 1 - \frac{6\sum d^2}{n(n^2 - 1)}...

$n = 10$ competitors. Competitor 1st Judge ($R1$) 2nd Judge ($R2$) 3rd Judge ($R3$) :---::---::---::---: 1 2 4 5 2 1 3 8 3 4 2 4 4 6 5 7 5 5 1 10 6 8 6 2 7 9 8 1 8 10 9 6 9 7 10 9 10 3 7 3 Formula: $$rs = 1 - \frac{6\sum d^2}{n(n^2 - 1)}...

125 marksTypes of samplingAnswer

What do you mean by sampling? Explain the difference between stratified sampling and cluster sampling. [5]

Sampling is the process of selecting a subset of individuals (called a sample) from a larger group (called the population) in order to draw conclusions or make inferences about the entire population. Since it is often impractical or impo...

135 marksApplication of Statistics in the field of Answer

State with suitable examples the role played by computer technology in applied statistics and the role of statistics in information technology. [5]

Computer technology has revolutionized the field of applied statistics by making complex calculations faster, more accurate, and more reliable. The key roles are described below: Computers can perform many statistical calculations easily...

Statistics I 2078 question paper

Methods of Measuring Dispersion and Comparison of Bursting Pressure

Given Data

Part 1: Methods of Measuring Dispersion

Part 2: Computation

Manufacturer A

Manufacturer B

Conclusion

Methods of Measuring Dispersion and Comparison of Bursting Pressure

Given Data

Part 1: Methods of Measuring Dispersion

Part 2: Computation

Manufacturer A

Manufacturer B

Conclusion

Joint PDF: f(x,y)=k(2x+3y)f(x,y) = k(2x + 3y)f(x,y)=k(2x+3y), 0≤x≤10 \le x \le 10≤x≤1, 0≤y≤10 \le y \le 10≤y≤1

Given data

(i) Constant k

(ii) Conditional PDF of X (given Y)

(iii) Independence

Joint PDF: $f(x,y) = k(2x + 3y)$, $0 \le x \le 1$, $0 \le y \le 1$

Given data

(i) Constant k

(ii) Conditional PDF of X (given Y)

(iii) Independence

Methods of Measuring Dispersion and Comparison of Bursting Pressure

Given Data

Part 1: Methods of Measuring Dispersion

Part 2: Computation

Manufacturer A

Manufacturer B

Conclusion

Joint PDF: f(x,y)=k(2x+3y)f(x,y) = k(2x + 3y)f(x,y)=k(2x+3y), 0≤x≤10 \le x \le 10≤x≤1, 0≤y≤10 \le y \le 10≤y≤1

Given data

(i) Constant k

(ii) Conditional PDF of X (given Y)

(iii) Independence

Joint PDF: $f(x,y) = k(2x + 3y)$ , $0 \le x \le 1$ , $0 \le y \le 1$

Joint PDF: $f(x,y) = k(2x + 3y)$ , $0 \le x \le 1$ , $0 \le y \le 1$