Maharashtra Board Class 12 Maths Part 2 Chapter 8 Probability Distributions 8.2 Solutions

Question 1. Check whether each of the following is p.d.f.
(1) \( f(x) = \begin{cases} x & \text{for } 0 \le x \le 1 \\ 2-x & \text{for } 1 < x < 2 \end{cases} \)
Solution:
Given function is
\( f(x) = x, 0 \le x \le 1 \)
Each \( f(x) \ge 0 \), as \( x \ge 0 \).
And \( \int_0^1 f(x)dx = \int_0^1 xdx \)
\( = \left[\frac{x^2}{2}\right]_0^1 \)
\( = \frac{1}{2} \)
Also, \( f(x) = 2-x, 1 \le x \le 2 \)
\( \implies \) Each \( f(x) \ge 0 \), as \( x \le 2 \).
And \( \int_1^2 f(x)dx = \int_1^2 (2-x)dx \)
\( = 2\int_1^2 1dx - \int_1^2 xdx \)
\( = 2[x]_1^2 - \left[\frac{x^2}{2}\right]_1^2 \)
\( = 2(2-1) - \frac{1}{2}(4-1) \)
\( = 2(1) - \frac{1}{2}(3) \)
\( = 2 - \frac{3}{2} \)
\( = \frac{4-3}{2} \)
\( = \frac{1}{2} \)
Now, for the total range of \( 0 \le x \le 2 \).
\( \int_0^2 f(x)dx = \int_0^1 f(x)dx + \int_1^2 f(x)dx \)
\( = \frac{1}{2} + \frac{1}{2} \)
\( = 1 \)
\( \therefore \) The given function is a p.d.f. of x.

(ii) \( f(x) = 2 \) for \( 0 < x < 1 \)
Solution:
Given function is
\( f(x) = 2 \) for \( 0 < x < 1 \) Each \( f(x) > 0 \),
But \( \int_0^1 f(x)dx = \int_0^1 2dx = 2[x]_0^1 \)
\( = 2(1) \)
\( = 2 > 1 \).
\( \therefore \) The given function is not a p.d.f.
In simple words: A probability density function (p.d.f.) must satisfy two conditions: it must be non-negative everywhere, and its total integral over the entire domain must equal 1. For the first case, both conditions are met, so it's a p.d.f. For the second case, the integral is 2, which is not 1, so it's not a p.d.f.

🎯 Exam Tip: Remember the two key conditions for a function to be a p.d.f.: \( f(x) \ge 0 \) for all x, and \( \int_{-\infty}^{\infty} f(x) dx = 1 \). Thoroughly checking both is crucial for full marks.

Question 2. The following is the p.d.f. of a r.v. X.
\( f(x) = \begin{cases} \frac{x}{8} & \text{for } 0 < x < 4 \\ 0 & \text{otherwise} \end{cases} \)
Find (i) P(X < 1.5), (ii) P(1 < X < 2), (iii) P(X > 2)
Solution:
\( f(x) = \begin{cases} \frac{x}{8} & \text{for } 0 < x < 4 \\ 0 & \text{otherwise} \end{cases} \)

(i) \( P(X < 1.5) = \int_0^{1.5} \frac{x}{8} dx \)
\( = \frac{1}{8} \int_0^{1.5} xdx \)
\( = \frac{1}{8} \left[\frac{x^2}{2}\right]_0^{1.5} \)
\( = \frac{1}{8 \times 2} [(1.5)^2 - 0^2] \)
\( = \frac{1}{16} [2.25 - 0] \)
\( = \frac{2.25}{16} \)
\( = 0.14 \)

(ii) \( P(1 < X < 2) = \int_1^2 \frac{x}{8} dx \)
\( = \frac{1}{8} \int_1^2 xdx \)
\( = \frac{1}{8} \left[\frac{x^2}{2}\right]_1^2 \)
\( = \frac{1}{16} [2^2 - 1^2] \)
\( = \frac{1}{16} [4 - 1] \)
\( = \frac{3}{16} \)
\( = 0.1875 \)

(iii) \( P(X > 2) = \int_2^4 \frac{x}{8} dx \)
\( = \frac{1}{8} \int_2^4 xdx \)
\( = \frac{1}{8} \left[\frac{x^2}{2}\right]_2^4 \)
\( = \frac{1}{16} [4^2 - 2^2] \)
\( = \frac{1}{16} [16 - 4] \)
\( = \frac{12}{16} \)
\( = \frac{3}{4} \)
\( = 0.75 \)
In simple words: This problem involves calculating probabilities for a continuous random variable given its probability density function (p.d.f.). We find the probability by integrating the p.d.f. over the specified interval for each part of the question.

🎯 Exam Tip: When calculating probabilities from a p.d.f., ensure you use the correct integration limits for each specific probability interval requested. Be careful with calculations involving decimals and fractions.

Question 3. It is felt that error in measurement of reaction temperature (in Celsius) in an experiment is a continuous r.v. with p.d.f.
\( f(x) = \begin{cases} \frac{x^3}{64} & \text{for } 0 \le x \le 4 \\ 0 & \text{otherwise} \end{cases} \)
(i) Verify whether f(x) is a p.d.f.
(ii) Find P(0 < X < 1).
(iii) Find the probability that X is between 1 and 3.
Solution:
(i) f(x) is p.d.f. of r.v. X if
(a) \( f(x) \ge 0, \forall x \in R \)
(b) \( \int_0^4 f(x)dx = 1 \)
i.e.(a) \( f(x) = \frac{x^3}{64} \), \( f(x) \ge 0, 0 \le x \le 4 \)

(b) \( \int_0^4 f(x)dx = \int_0^4 \frac{x^3}{64}dx \)
\( = \frac{1}{64} \left[\frac{x^4}{4}\right]_0^4 \)
\( = \frac{1}{256} [4^4 - 0^4] \)
\( = \frac{256}{256} \)
\( = 1 \)
Hence, f(x) is a p.d.f. of r.v. X

(ii) \( P(0 < X \le 1) = \int_0^1 \frac{x^3}{64} dx \)
\( = \frac{1}{64} \left[\frac{x^4}{4}\right]_0^1 \)
\( = \frac{1}{256} [1^4 - 0] \)
\( = \frac{1}{256} \)

(iii) \( P(1 < X < 3) = \int_1^3 \frac{x^3}{64} dx \)
\( = \frac{1}{64} \left[\frac{x^4}{4}\right]_1^3 \)
\( = \frac{1}{256} [3^4 - 1^4] \)
\( = \frac{81 - 1}{256} \)
\( = \frac{80}{256} \)
\( = \frac{5}{16} \)
In simple words: First, we verified that the given function is indeed a probability density function by checking its non-negativity and confirming that its integral over the entire range is 1. Then, we calculated the probabilities for specific intervals by integrating the p.d.f. over those ranges.

🎯 Exam Tip: When verifying a p.d.f., ensure both conditions (\( f(x) \ge 0 \) and total integral equals 1) are explicitly checked. For probability calculations, correct limits of integration are crucial.

Question 4. Find k, if the following function represents the p.d.f. of a r.v. X.
(i) \( f(x) = \begin{cases} kx & \text{for } 0 < x < 2 \\ 0 & \text{otherwise} \end{cases} \)
Also find \( P[\frac{1}{4} < X < \frac{1}{2}] \)
Solution:
\( f(x) = \begin{cases} kx & \text{for } 0 < x < 2, \\ 0 & \text{otherwise} \end{cases} \)
\( \therefore f(x) = kx, 0 < x < 2 \)
We know that for a p.d.f., \( \int_0^2 f(x)dx = 1 \)
\( \implies \int_0^2 kx dx = 1 \)
\( \implies k \int_0^2 xdx = 1 \)
\( \implies k \left[\frac{x^2}{2}\right]_0^2 = 1 \)
\( \implies k \left(\frac{2^2 - 0^2}{2}\right) = 1 \)
\( \implies k \left(\frac{4}{2}\right) = 1 \)
\( \implies 2k = 1 \)
\( \implies k = \frac{1}{2} \)
\( \therefore f(x) = kx = \frac{x}{2} \)

\( P[\frac{1}{4} < X < \frac{3}{2}] = \int_{\frac{1}{4}}^{\frac{3}{2}} \frac{x}{2} dx \)
\( = \frac{1}{2} \int_{\frac{1}{4}}^{\frac{3}{2}} xdx \)
\( = \frac{1}{2} \left[\frac{x^2}{2}\right]_{\frac{1}{4}}^{\frac{3}{2}} \)
\( = \frac{1}{4} \left[\left(\frac{3}{2}\right)^2 - \left(\frac{1}{4}\right)^2\right] \)
\( = \frac{1}{4} \left[\frac{9}{4} - \frac{1}{16}\right] \)
\( = \frac{1}{4} \left[\frac{36 - 1}{16}\right] \)
\( = \frac{1}{4} \left[\frac{35}{16}\right] \)
\( = \frac{35}{64} \)
\( = 0.55 \)

(ii) \( f(x) = \begin{cases} kx(1-x) & \text{for } 0 < x < 1 \\ 0 & \text{otherwise} \end{cases} \)
Also find (a) \( P[\frac{1}{4} < X < \frac{1}{2}] \), (b) \( P[X < \frac{1}{2}] \)
Solution:
We know that \( \int_0^1 f(x)dx = 1 \)
\( \implies \int_0^1 kx(1-x)dx = 1 \)
\( \implies k \int_0^1 (x - x^2)dx = 1 \)
\( \implies k \left[\frac{x^2}{2} - \frac{x^3}{3}\right]_0^1 = 1 \)
\( \implies k \left[\left(\frac{1^2}{2} - \frac{1^3}{3}\right) - \left(\frac{0^2}{2} - \frac{0^3}{3}\right)\right] = 1 \)
\( \implies k \left[\frac{1}{2} - \frac{1}{3}\right] = 1 \)
\( \implies k \left[\frac{3 - 2}{6}\right] = 1 \)
\( \implies k \left[\frac{1}{6}\right] = 1 \)
\( \implies k = 6 \)
\( \therefore f(x) = 6x(1-x) \)

(a) \( P[\frac{1}{4} < X < \frac{1}{2}] = \int_{\frac{1}{4}}^{\frac{1}{2}} 6x(1-x)dx = 6 \int_{\frac{1}{4}}^{\frac{1}{2}} (x - x^2)dx \)
\( = 6 \left[\frac{x^2}{2} - \frac{x^3}{3}\right]_{\frac{1}{4}}^{\frac{1}{2}} \)
\( = 6 \left[\left(\frac{(\frac{1}{2})^2}{2} - \frac{(\frac{1}{2})^3}{3}\right) - \left(\frac{(\frac{1}{4})^2}{2} - \frac{(\frac{1}{4})^3}{3}\right)\right] \)
\( = 6 \left[\left(\frac{1}{8} - \frac{1}{24}\right) - \left(\frac{1}{32} - \frac{1}{192}\right)\right] \)
\( = 6 \left[\left(\frac{3-1}{24}\right) - \left(\frac{6-1}{192}\right)\right] \)
\( = 6 \left[\frac{2}{24} - \frac{5}{192}\right] \)
\( = 6 \left[\frac{1}{12} - \frac{5}{192}\right] \)
\( = 6 \left[\frac{16 - 5}{192}\right] \)
\( = 6 \left[\frac{11}{192}\right] \)
\( = \frac{66}{192} \)
\( = \frac{11}{32} \)
\( = 0.34375 \)

(b) \( P[X < \frac{1}{2}] = \int_0^{\frac{1}{2}} 6x(1-x)dx = 6 \int_0^{\frac{1}{2}} (x - x^2)dx \)
\( = 6 \left[\frac{x^2}{2} - \frac{x^3}{3}\right]_0^{\frac{1}{2}} \)
\( = 6 \left[\left(\frac{(\frac{1}{2})^2}{2} - \frac{(\frac{1}{2})^3}{3}\right) - (0 - 0)\right] \)
\( = 6 \left[\frac{1}{8} - \frac{1}{24}\right] \)
\( = 6 \left[\frac{3 - 1}{24}\right] \)
\( = 6 \left[\frac{2}{24}\right] \)
\( = 6 \left[\frac{1}{12}\right] \)
\( = \frac{6}{12} \)
\( = \frac{1}{2} \)
\( = 0.5 \)
In simple words: For each case, we first used the property that the total probability for a p.d.f. must be 1 to find the constant 'k' by integrating the function over its defined range and setting the result equal to 1. Once 'k' was found, we calculated the specific probabilities for the given intervals by integrating the updated p.d.f. over those new limits.

🎯 Exam Tip: When solving for 'k', remember the integral of the p.d.f. over its entire domain must always equal 1. Pay close attention to the limits of integration for each probability calculation, especially with fractional bounds.

Question 5. Let X be the amount of time for which a book is taken out of the library by a randomly selected student and suppose that X has p.d.f.
\( f(x) = \begin{cases} 0.5x & \text{for } 0 \le x \le 2 \\ 0 & \text{otherwise} \end{cases} \)
Calculate (i) P(X < 1), (ii) P(0.5 < X < 1.5), (iii) P(X < 1.5).
Solution:
Given p.d.f. of X is \( f(x) = 0.5x \) for \( 0 \le x \le 2 \)
\( \therefore \) Its c.d.f. F(x) is given by
\( F(x) = \int_0^x f(y)dy \), where \( f(y) = 0.5y \)
\( = \int_0^x 0.5y dy \)
\( = 0.5 \left[\frac{y^2}{2}\right]_0^x \)
\( = 0.5 \times \frac{x^2}{2} \)
\( = 0.25x^2 \)

(i) \( P(X < 1) = F(1) \)
\( = 0.25(1)^2 \)
\( = 0.25 \)

(ii) \( P(0.5 < X < 1.5) = F(1.5) - F(0.5) \)
\( = 0.25(1.5)^2 - 0.25(0.5)^2 \)
\( = 0.25[2.25 - 0.25] \)
\( = 0.25(2) \)
\( = 0.5 \)

(iii) \( P(X \ge 1.5) = 1 - P(X < 1.5) \)
\( = 1 - F(1.5) \)
\( = 1 - 0.25(1.5)^2 \)
\( = 1 - 0.25(2.25) \)
\( = 1 - 0.5625 \)
\( = 0.4375 \)
In simple words: This problem involves using the cumulative distribution function (c.d.f.) to find probabilities. First, we derived the c.d.f. by integrating the given p.d.f. From the c.d.f., we directly calculated the probabilities for specific time intervals, using the property \( P(a < X < b) = F(b) - F(a) \) and \( P(X \ge a) = 1 - P(X < a) = 1 - F(a) \).

🎯 Exam Tip: Deriving the correct c.d.f. \( F(x) \) from the p.d.f. \( f(x) \) is the first critical step. Remember that \( P(X < a) = F(a) \), \( P(a < X < b) = F(b) - F(a) \), and \( P(X \ge a) = 1 - F(a) \) for continuous distributions.

Question 6. Suppose X is the waiting time (in minutes) for a bus and its p.d.f. is given by
\( f(x) = \begin{cases} \frac{1}{5} & \text{for } 0 \le x \le 5 \\ 0 & \text{otherwise} \end{cases} \)
Find the probability that (i) waiting time is between 1 and 3 minutes, (ii) waiting time is more than 4 minutes.
Solution:
p.d.f. of r.v. X is given by
\( f(x) = \frac{1}{5} \) for \( 0 \le x \le 5 \)
This is a constant function.

(i) Probability that waiting time X is between 1 and 3 minutes
i.e. \( P(1 < X < 3) = \int_1^3 f(x) dx \)
\( = \int_1^3 \frac{1}{5} dx \)
\( = \frac{1}{5} [x]_1^3 \)
\( = \frac{1}{5} [3 - 1] \)
\( = \frac{2}{5} \)
\( = 0.4 \)

(ii) Probability that waiting time X is more than 4 minutes
i.e. \( P(X > 4) = \int_4^5 f(x) dx \)
\( = \int_4^5 \frac{1}{5} dx \)
\( = \frac{1}{5} [x]_4^5 \)
\( = \frac{1}{5} [5 - 4] \)
\( = \frac{1}{5} \)
\( = 0.2 \)
In simple words: For a constant probability density function, calculating probabilities for given intervals simply involves integrating the constant over those intervals. This is equivalent to finding the area of a rectangle formed by the constant height of the p.d.f. and the width of the interval.

🎯 Exam Tip: When the p.d.f. is a constant over its defined range (a uniform distribution), probabilities can often be found by calculating the length of the interval multiplied by the constant p.d.f. value, ensuring the interval is within the valid range.

Question 7. Suppose error involved in making a certain measurement is a continuous r.v. X with p.d.f.
\( f(x) = \begin{cases} k(4 - x^2) & \text{for } -2 \le x \le 2 \\ 0 & \text{otherwise} \end{cases} \)
Compute (i) P(X > 0), (ii) P(-1 < X < 1), (iii) P(X < -0.5 or X > 0.5)
Solution:
Since given f(x) is a p.d.f. of r.v. X
Since \( -2 \le x \le 2 \)
\( \therefore x^2 \le 4 \)
\( \therefore 4 - x^2 \ge 0 \)
\( \therefore k(4 - x^2) \ge 0 \)
\( \therefore k \ge 0 \text{ [: f(x) \ge 0]} \)
Also \( \int_{-2}^2 f(x) dx = 1 \)
\( \implies \int_{-2}^2 k(4-x^2) dx = 1 \)
\( \implies 2k \int_0^2 (4-x^2) dx = 1 \) (\( \therefore 4-x^2 \) is an even function)
\( \implies 2k \left[4x - \frac{x^3}{3}\right]_0^2 = 1 \)
\( \implies 2k \left[4(2) - \frac{2^3}{3}\right] - 0 = 1 \)
\( \implies 2k \left[8 - \frac{8}{3}\right] = 1 \)
\( \implies 2k \left[\frac{24 - 8}{3}\right] = 1 \)
\( \implies 2k \left[\frac{16}{3}\right] = 1 \)
\( \implies \frac{32k}{3} = 1 \)
\( \implies k = \frac{3}{32} \)
\( \therefore \) p.d.f. of X is
\( f(x) = \frac{3}{32} (4-x^2) \) for \( -2 \le x \le 2 \)

(i) \( P(X > 0) = \int_0^2 f(x)dx \)
\( = \int_0^2 \frac{3}{32} (4-x^2) dx \)
\( = \frac{3}{32} \int_0^2 (4-x^2) dx \)
\( = \frac{3}{32} \left[4x - \frac{x^3}{3}\right]_0^2 \)
\( = \frac{3}{32} \left[4(2) - \frac{2^3}{3}\right] - 0 \)
\( = \frac{3}{32} \left[8 - \frac{8}{3}\right] \)
\( = \frac{3}{32} \left[\frac{24 - 8}{3}\right] \)
\( = \frac{3}{32} \left[\frac{16}{3}\right] \)
\( = \frac{16}{32} \)
\( = \frac{1}{2} \)
\( = 0.5 \)

(ii) \( P(-1 < X < 1) = \int_{-1}^1 f(x)dx \)
\( = \int_{-1}^1 \frac{3}{32} (4-x^2) dx \)
\( = \frac{3}{32} \int_{-1}^1 (4-x^2) dx \)
\( = \frac{3}{32} \times 2 \int_0^1 (4-x^2) dx \) (\( \because \) even function)
\( = \frac{3}{16} \left[4x - \frac{x^3}{3}\right]_0^1 \)
\( = \frac{3}{16} \left[4(1) - \frac{1^3}{3}\right] - 0 \)
\( = \frac{3}{16} \left[4 - \frac{1}{3}\right] \)
\( = \frac{3}{16} \left[\frac{12 - 1}{3}\right] \)
\( = \frac{3}{16} \left[\frac{11}{3}\right] \)
\( = \frac{11}{16} \)
\( = 0.6875 \)

(iii) \( P(X < -0.5 \text{ or } X > 0.5) \)
\( = \int_{-2}^{-0.5} f(x)dx + \int_{0.5}^2 f(x)dx \)
\( = \int_{-2}^{-0.5} \frac{3}{32} (4-x^2) dx + \int_{0.5}^2 \frac{3}{32} (4-x^2) dx \)
Since \( 4 - x^2 \) is a symmetric function
Area under the curve between -2 and -0.5
and 0.5 and 2 are equal.
\( = 2 \int_{0.5}^2 \frac{3}{32} (4-x^2) dx \)
\( = \frac{3}{16} \int_{0.5}^2 (4-x^2) dx \)
\( = \frac{3}{16} \left[4x - \frac{x^3}{3}\right]_{0.5}^2 \)
\( = \frac{3}{16} \left[\left(4(2) - \frac{2^3}{3}\right) - \left(4(0.5) - \frac{(0.5)^3}{3}\right)\right] \)
\( = \frac{3}{16} \left[\left(8 - \frac{8}{3}\right) - \left(2 - \frac{0.125}{3}\right)\right] \)
\( = \frac{3}{16} \left[\left(\frac{24-8}{3}\right) - \left(\frac{6-0.125}{3}\right)\right] \)
\( = \frac{3}{16} \left[\frac{16}{3} - \frac{5.875}{3}\right] \)
\( = \frac{3}{16} \left[\frac{16 - 5.875}{3}\right] \)
\( = \frac{3}{16} \left[\frac{10.125}{3}\right] \)
\( = \frac{10.125}{16} \)
\( = 0.6328 \)
In simple words: First, we determined the value of 'k' by using the fundamental property that the integral of a probability density function over its entire domain must be 1. Then, we calculated the probabilities for specific intervals by integrating the determined p.d.f. over the given limits, leveraging the symmetry of the function to simplify calculations where applicable.

🎯 Exam Tip: Always exploit symmetry in p.d.f.s (like even functions) to simplify calculations, especially when dealing with symmetric intervals around zero. Remember that \( \int_{-a}^a f(x)dx = 2 \int_0^a f(x)dx \) if \( f(x) \) is an even function.

Question 8. Following is the p.d.f. of a continuous r.v. X.
\( f(x) = \begin{cases} \frac{x}{8} & \text{for } 0 < x < 4 \\ 0 & \text{otherwise} \end{cases} \)
(i) Find an expression for the c.d.f. of X.
(ii) Find F(x) at x = 0.5, 1.7, and 5.
Solution:
The p.d.f. of a continuous r.v. X is
\( f(x) = \begin{cases} \frac{x}{8} & \text{for } 0 < x < 4 \\ 0 & \text{otherwise} \end{cases} \)

(i) c.d.f. of continuous r.v. X is given by
\( F(x) = \int_{-\infty}^x f(y)dy \)
\( \therefore F(x) = \int_0^x \frac{y}{8}dy \)
\( = \frac{1}{8} \int_0^x ydy \)
\( = \frac{1}{8} \left[\frac{y^2}{2}\right]_0^x \)
\( = \frac{1}{16} [x^2 - 0^2] \)
\( = \frac{x^2}{16} \)

(ii) \( F(0.5) = \frac{(0.5)^2}{16} = \frac{0.25}{16} = \frac{1}{64} = 0.015 \)
\( F(1.7) = \frac{(1.7)^2}{16} = \frac{2.89}{16} = 0.18 \)
For any of x greater than or equal to 4, F(x) = 1
\( \therefore F(5) = 1 \)
In simple words: First, we derived the cumulative distribution function (c.d.f.) by integrating the given probability density function (p.d.f.) from the lower bound of the domain up to x. Then, we used this derived c.d.f. to find the cumulative probabilities at specific points (x = 0.5, 1.7, and 5), remembering that F(x) becomes 1 for any x beyond the p.d.f.'s upper limit.

🎯 Exam Tip: When finding the c.d.f. \( F(x) \), ensure you define it piecewise for all possible x values, accounting for \( F(x) = 0 \) for x below the domain, the integral for x within the domain, and \( F(x) = 1 \) for x above the domain. This complete definition is crucial for evaluating F(x) at any point.

Question 9. The p.d.f. of a continuous r.v. X is
\( f(x) = \begin{cases} \frac{3x^2}{8} & \text{for } 0 < x < 2 \\ 0 & \text{otherwise} \end{cases} \)
Determine the c.d.f. of X and hence find (i) P(X < 1), (ii) P(X < -2), (iii) P(X > 0), (iv) P(1 < X < 2).
Solution:
The p.d.f. of a continuous r.v. X is
\( f(x) = \begin{cases} \frac{3x^2}{8} & \text{for } 0 < x < 2, \\ 0 & \text{otherwise} \end{cases} \)
c.d.f. of X is given by
\( F(x) = \int_0^x f(y)dy \)
\( = \int_0^x \frac{3y^2}{8} dy \)
\( = \frac{3}{8} \left[\frac{y^3}{3}\right]_0^x \)
\( = \frac{1}{8} [x^3 - 0^3] \)
\( = \frac{x^3}{8} \)

(i) \( P(X < 1) = F(1) = \frac{1^3}{8} = \frac{1}{8} \)

(ii) \( P(X < -2) = 0 \) \( \because \) Range of X is (0, 2)

(iii) \( P(X > 0) = 1 - P(X \le 0) \)
\( = 1 - F(0) \)
\( = 1 - \left[\frac{0^3}{8}\right] \)
\( = 1 - 0 \)
\( = 1 \)

(iv) \( P(1 < X < 2) = F(2) - F(1) \)
\( = \frac{2^3}{8} - \frac{1^3}{8} \)
\( = \frac{8}{8} - \frac{1}{8} \)
\( = \frac{8 - 1}{8} \)
\( = \frac{7}{8} \)
In simple words: We first found the cumulative distribution function (c.d.f.) by integrating the given p.d.f. The c.d.f. then allowed us to directly calculate probabilities for various intervals, noting that probabilities outside the defined range of the random variable are zero and the total probability over the entire range is one.

🎯 Exam Tip: When asked to determine the c.d.f., remember to integrate the p.d.f. with respect to a dummy variable (e.g., y) from the lower limit of the support to 'x'. For probabilities, always refer to the c.d.f.'s properties and the defined range of the random variable.

Question 10. If a r.v. X has p.d.f.
\( f(x) = \begin{cases} \frac{c}{x} & \text{for } 1 < x < 3, c > 0 \\ 0 & \text{otherwise} \end{cases} \)
Find c, E(X) and V(X). Also find f(x).
Solution:
The p.d.f. of r.v. X is
\( f(x) = \frac{c}{x}, 1 < x < 3, c > 0 \)
In simple words: This question requires finding the constant 'c', the expected value (mean), and the variance of a continuous random variable given its probability density function, followed by specifying the complete function itself.

🎯 Exam Tip: To find the constant 'c' for a p.d.f., always integrate the function over its entire range and set the result equal to 1. This is the foundational step for all subsequent calculations involving expected value and variance. Ensure correct integration techniques, especially for logarithmic functions in this case, are applied precisely.

Question 10. If a r.v. X has p.d.f. \( f(x) = \begin{cases} \frac{c}{x} & \text{for } 1 < x < 3, c > 0 \\ 0 & \text{otherwise} \end{cases} \) Find c, E(X) and V(X). Also find f(x).
Answer: Solution: The p.d.f. of r.v. X is \( f(x) = \frac{c}{x}, 1 < x < 3, c > 0 \) For p.d.f. of X, we have \( \int_{-\infty}^{\infty} f(x)dx = 1 \)
\( \implies \int_1^3 \frac{c}{x} dx = 1 \)
\( \implies c[\log x]_1^3 = 1 \)
\( \implies c[\log 3 - \log 1] = 1 \)
\( \implies c \log \left(\frac{3}{1}\right) = 1 \)
\( \implies c \log 3 = 1 \)
\( \implies c = \frac{1}{\log 3} \)
\( \therefore f(x) = \frac{1}{x \log 3} \)
Now, let's find E(X):
\( E(X) = \int_1^3 xf(x)dx \)
\( \implies E(X) = \int_1^3 x \cdot \frac{1}{x \log 3} dx \)
\( \implies E(X) = \int_1^3 \frac{1}{\log 3} dx \)
\( \implies E(X) = \frac{1}{\log 3} [x]_1^3 \)
\( \implies E(X) = \frac{1}{\log 3} (3-1) \)
\( \implies E(X) = \frac{2}{\log 3} \)
Hence, \( E(X) = \frac{2}{\log 3} \)
Next, let's find E(X²):
\( E(X^2) = \int_1^3 x^2 \cdot f(x)dx \)
\( \implies E(X^2) = \int_1^3 x^2 \cdot \frac{1}{x \log 3}dx \)
\( \implies E(X^2) = \frac{1}{\log 3} \int_1^3 x dx \)
\( \implies E(X^2) = \frac{1}{\log 3} \left[\frac{x^2}{2}\right]_1^3 \)
\( \implies E(X^2) = \frac{1}{\log 3} \left[\frac{3^2}{2} - \frac{1^2}{2}\right] \)
\( \implies E(X^2) = \frac{1}{\log 3} \left[\frac{9}{2} - \frac{1}{2}\right] \)
\( \implies E(X^2) = \frac{1}{\log 3} \cdot \frac{8}{2} \)
\( \implies E(X^2) = \frac{4}{\log 3} \)
Finally, let's find V(X):
\( \therefore V(X) = E(X^2) - [E(X)]^2 \)
\( \implies V(X) = \frac{4}{\log 3} - \left(\frac{2}{\log 3}\right)^2 \)
\( \implies V(X) = \frac{4}{\log 3} - \frac{4}{(\log 3)^2} \)
The c.d.f. of F(x) is given by:
\( F(x) = \int_1^x f(y)dy \)
\( \implies F(x) = \int_1^x \frac{1}{y \log 3} dy \)
\( \implies F(x) = \frac{1}{\log 3} \int_1^x \frac{1}{y} dy \)
\( \implies F(x) = \frac{1}{\log 3} [\log y]_1^x \)
\( \implies F(x) = \frac{1}{\log 3} [\log x - \log 1] \)
\( \implies F(x) = \frac{\log x}{\log 3} \)
In simple words: First, we use the property that the total probability is 1 to find the constant 'c'. Then, we calculate the expected value E(X) and E(X²) by integrating x*f(x) and x²*f(x) respectively over the given range. Finally, the variance V(X) is found using E(X²) and E(X). The cumulative distribution function F(x) is obtained by integrating f(y) from 1 to x.

🎯 Exam Tip: Remember to correctly identify the limits of integration for each calculation (c, E(X), E(X²), V(X), and F(x)). Pay close attention to the definition of f(x) for different ranges of x.