Samacheer Kalvi Class 12 Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 1.2

 

Question 1. Construct cumulative distribution function for the given probability distribution.

Answer: The cumulative distribution function (CDF), denoted \( F(x) \), gives the probability that a random variable \( X \) takes a value less than or equal to a specific value \( x \). We calculate \( F(x) \) for each given \( x \) by summing the probabilities up to that point. \( F(0) = P(X=0) = 0.3 \) \( F(1) = P(X \le 1) = P(X=0) + P(X=1) = 0.3 + 0.2 = 0.5 \) \( F(2) = P(X \le 2) = P(X=0) + P(X=1) + P(X=2) = 0.3 + 0.2 + 0.4 = 0.9 \) \( F(3) = P(X \le 3) = P(X=0) + P(X=1) + P(X=2) + P(X=3) = 0.3 + 0.2 + 0.4 + 0.1 = 1 \) The CDF always starts at 0 and goes up to 1, showing how the probability accumulates.
In simple words: To get the cumulative distribution, we keep adding the probabilities for each value of X. For example, for \(F(1)\), we add \(P(X=0)\) and \(P(X=1)\). The last value of the cumulative function should always be 1.

🎯 Exam Tip: Remember that \(F(x)\) must be non-decreasing, and \(F(-\infty) = 0\) while \(F(\infty) = 1\). This is a good way to check your calculations.

 

Question 2. Let X be a discrete random variable with the following p.m.f Find and plot the c.d.f. of X.\[ P(x)= \begin{cases} 0.3 & \text{for } x=3 \\ 0.2 & \text{for } x=5 \\ 0.3 & \text{for } x=8 \\ 0.2 & \text{for } x=10 \\ 0 & \text{otherwise} \end{cases} \]Answer: We first find the cumulative distribution function (CDF), \( F(x) \), by summing the probabilities for \(X\) up to a certain point. \( F(3) = P(X \le 3) = P(X=3) = 0.3 \) \( F(5) = P(X \le 5) = P(X=3) + P(X=5) = 0.3 + 0.2 = 0.5 \) \( F(8) = P(X \le 8) = P(X=3) + P(X=5) + P(X=8) = 0.3 + 0.2 + 0.3 = 0.8 \) \( F(10) = P(X \le 10) = P(X=3) + P(X=5) + P(X=8) + P(X=10) = 0.3 + 0.2 + 0.3 + 0.2 = 1 \) Thus, the cumulative distribution function for \( X \) is: \[ F(x)= \begin{cases} 0 & \text{for } x < 3 \\ 0.3 & \text{for } 3 \le x < 5 \\ 0.5 & \text{for } 5 \le x < 8 \\ 0.8 & \text{for } 8 \le x < 10 \\ 1 & \text{for } x \ge 10 \end{cases} \] The plot of the CDF is a step function, as shown below. The value of \( F(x) \) jumps at each discrete point where \( X \) has a positive probability, and stays constant between these points.
In simple words: First, we add up the probabilities to find the cumulative distribution function for each specific value. Then, we draw a graph where the line stays flat between the values, and then jumps up at each new value, like steps on a staircase. The graph should start at zero and end at one.

🎯 Exam Tip: When plotting a CDF for a discrete random variable, always show open circles at the value just before a jump and closed circles at the value where the jump occurs to clearly indicate the intervals.

 

Question 3. The discrete random variable X has the following probability function. Show that \( k = \frac { 1 }{18} \).\[ P(X=x)= \begin{cases} kx & \text{for } x=2,4,6 \\ k(x-2) & \text{for } x=8 \\ 0 & \text{otherwise} \end{cases} \]Answer: For a probability mass function (PMF), the sum of all probabilities for all possible values of \(X\) must be equal to 1. First, we find the probabilities for each value of \(X\): \( P(X=2) = k(2) = 2k \) \( P(X=4) = k(4) = 4k \) \( P(X=6) = k(6) = 6k \) \( P(X=8) = k(8-2) = k(6) = 6k \) Now, we sum these probabilities and set the total equal to 1: \( \sum P(X=x) = P(X=2) + P(X=4) + P(X=6) + P(X=8) = 1 \) \( 2k + 4k + 6k + 6k = 1 \) \( 18k = 1 \) Next, we solve for \(k\): \( k = \frac{1}{18} \) This confirms the given value of \(k\).
In simple words: For any probability function, if you add up all the chances (probabilities) for every possible outcome, the total must always be 1. We used this rule to find the value of \(k\).

🎯 Exam Tip: Always remember that the sum of probabilities for a valid PMF must be exactly 1. This property is crucial for finding unknown constants like \(k\).

 

Question 4. The discrete random variable X has the probability function.

Answer: For any probability mass function (PMF), the sum of all probabilities for all possible values of \(X\) must equal 1. From the given table, we have the probabilities for \(X=1, 2, 3, 4\): \( P(X=1) = k \) \( P(X=2) = 2k \) \( P(X=3) = 3k \) \( P(X=4) = 4k \) Now, we add these probabilities together and set the sum to 1: \( \sum P(X=x) = P(X=1) + P(X=2) + P(X=3) + P(X=4) = 1 \) \( k + 2k + 3k + 4k = 1 \) \( 10k = 1 \) Finally, we solve for \(k\): \( k = \frac{1}{10} = 0.1 \) This value ensures that the total probability is 1.
In simple words: We need to find the value of \(k\). Since all probabilities must add up to 1, we add all the \(k\) terms from the table and set the total to 1. Then we just solve for \(k\).

🎯 Exam Tip: Always sum the probabilities to 1 to find an unknown constant. This is a fundamental property of probability distributions, whether discrete or continuous.

 

Question 5. Two coins are tossed simultaneously. Getting a head is termed as success. Find the probability distribution of the number of successes.

Answer: When two coins are tossed at the same time, the possible outcomes are HH (Head-Head), HT (Head-Tail), TH (Tail-Head), and TT (Tail-Tail). Each of these outcomes has an equal probability of \( \frac{1}{4} \). Let \( X \) be the random variable representing the number of heads (successes). If the outcome is HH, \( X=2 \) (2 heads). If the outcome is HT, \( X=1 \) (1 head). If the outcome is TH, \( X=1 \) (1 head). If the outcome is TT, \( X=0 \) (0 heads). The possible values for \( X \) are 0, 1, and 2. Now we find the probability for each value of \( X \): \( P(X=0) = P(\text{TT}) = \frac{1}{4} \) \( P(X=1) = P(\text{HT}) + P(\text{TH}) = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} \) \( P(X=2) = P(\text{HH}) = \frac{1}{4} \) The probability distribution of the number of successes is:This table shows how the probability is spread across the different numbers of heads.
In simple words: When you flip two coins, you can get 0, 1, or 2 heads. We list all possible ways these heads can appear (HH, HT, TH, TT) and then count how many times each number of heads shows up. Then we write down the chance for each number of heads in a small table.

🎯 Exam Tip: For coin toss problems, always list all possible outcomes in the sample space first. This helps ensure that you correctly assign probabilities to each value of the random variable.

 

Question 6. The discrete random variable X has the probability function.

Answer:
(i) To find \(k\), we use the property that the sum of all probabilities in a probability mass function (PMF) must equal 1. \( \sum P(x) = P(0) + P(1) + P(2) + P(3) + P(4) + P(5) + P(6) + P(7) = 1 \) Substitute the given expressions for \( P(x) \): \( 0 + k + 2k + 2k + 3k + k^2 + 2k^2 + (7k^2 + k) = 1 \) Combine like terms: \( (k+2k+2k+3k+k) + (k^2+2k^2+7k^2) = 1 \) \( 9k + 10k^2 = 1 \) Rearrange into a standard quadratic equation: \( 10k^2 + 9k - 1 = 0 \) We can factor this quadratic equation: \( 10k^2 + 10k - k - 1 = 0 \) \( 10k(k+1) - 1(k+1) = 0 \) \( (k+1)(10k-1) = 0 \) This gives two possible values for \(k\): \( k+1=0 \implies k = -1 \) \( 10k-1=0 \implies k = \frac{1}{10} \) Since probability values cannot be negative, \(k=-1\) is not a valid solution. Therefore, the value of \(k\) is \( \frac{1}{10} \).
(ii) Now we evaluate the probabilities: \( P(x < 6) \), \( P(x \ge 6) \) and \( P(0 < x < 5) \). First, substitute \( k = \frac{1}{10} \) into the expressions for \(P(x)\) from the table: \( P(1) = \frac{1}{10} \) \( P(2) = \frac{2}{10} \) \( P(3) = \frac{2}{10} \) \( P(4) = \frac{3}{10} \) \( P(5) = \left(\frac{1}{10}\right)^2 = \frac{1}{100} \) \( P(6) = 2\left(\frac{1}{10}\right)^2 = \frac{2}{100} \) \( P(7) = 7\left(\frac{1}{10}\right)^2 + \frac{1}{10} = \frac{7}{100} + \frac{10}{100} = \frac{17}{100} \) Now, calculate the required probabilities: \( P(x < 6) = P(0) + P(1) + P(2) + P(3) + P(4) + P(5) \) \( = 0 + \frac{1}{10} + \frac{2}{10} + \frac{2}{10} + \frac{3}{10} + \frac{1}{100} \) \( = \frac{8}{10} + \frac{1}{100} = \frac{80}{100} + \frac{1}{100} = \frac{81}{100} \) \( P(x \ge 6) = P(6) + P(7) \) \( = \frac{2}{100} + \frac{17}{100} = \frac{19}{100} \) \( P(0 < x < 5) = P(1) + P(2) + P(3) + P(4) \) \( = \frac{1}{10} + \frac{2}{10} + \frac{2}{10} + \frac{3}{10} = \frac{8}{10} \)
(iii) We need to find the minimum value of \(x\) for which \( P(X \le x) > \frac{1}{2} \). We calculate the cumulative probabilities: \( P(X \le 0) = P(0) = 0 \) (Not greater than 0.5) \( P(X \le 1) = P(0) + P(1) = 0 + \frac{1}{10} = \frac{1}{10} = 0.1 \) (Not greater than 0.5) \( P(X \le 2) = P(X \le 1) + P(2) = 0.1 + \frac{2}{10} = 0.1 + 0.2 = 0.3 \) (Not greater than 0.5) \( P(X \le 3) = P(X \le 2) + P(3) = 0.3 + \frac{2}{10} = 0.3 + 0.2 = 0.5 \) (Not greater than 0.5, it is equal to 0.5) \( P(X \le 4) = P(X \le 3) + P(4) = 0.5 + \frac{3}{10} = 0.5 + 0.3 = 0.8 \) (Greater than 0.5) The minimum value of \(x\) for which \( P(X \le x) > \frac{1}{2} \) is 4.
In simple words: (i) To find \(k\), we add up all the parts that have \(k\) and \(k^2\) in them. This total sum must always be 1 for a probability function. We then solve the equation to find \(k\). Remember that \(k\) cannot be a negative number. (ii) Once we know \(k\), we plug its value back into the probabilities to find the chance of \(X\) being less than 6, or 6 or more, or between 0 and 5. (iii) For the last part, we keep adding probabilities from the start until the total is more than 0.5. The first \(x\) value where this happens is our answer.

🎯 Exam Tip: Always double-check your calculations when solving for \(k\) in quadratic equations to ensure you select the valid non-negative solution. For cumulative probabilities, add up probabilities carefully step-by-step.

 

Question 7. The distribution of a continuous random variable X in range (-3, 3) is given by p.d.f. Verify that the area under the curve is unity.\[ f(x)= \begin{cases} \frac{1}{16}(3+x)^2 & \text{for } -3 \le x \le -1 \\ \frac{1}{16}(6-2x^2) & \text{for } -1 \le x \le 1 \\ \frac{1}{16}(3-x)^2 & \text{for } 1 \le x \le 3 \end{cases} \]Answer: For \( f(x) \) to be a valid probability density function (PDF), the total area under its curve must be equal to 1. We verify this by integrating \( f(x) \) over its entire range, which is from -3 to 3. The integral is split into three parts based on the definition of \( f(x) \): \( A = \int_{-3}^{3} f(x)dx \) \( A = \int_{-3}^{-1} \frac{1}{16}(3+x)^2 dx + \int_{-1}^{1} \frac{1}{16}(6-2x^2) dx + \int_{1}^{3} \frac{1}{16}(3-x)^2 dx \) We can factor out \( \frac{1}{16} \) from the integral: \( A = \frac{1}{16} \left[ \int_{-3}^{-1} (3+x)^2 dx + \int_{-1}^{1} (6-2x^2) dx + \int_{1}^{3} (3-x)^2 dx \right] \) Now, we calculate each integral: 1. \( \int_{-3}^{-1} (3+x)^2 dx = \left[ \frac{(3+x)^3}{3} \right]_{-3}^{-1} = \frac{(3-1)^3}{3} - \frac{(3-3)^3}{3} = \frac{2^3}{3} - 0 = \frac{8}{3} \) 2. \( \int_{-1}^{1} (6-2x^2) dx = \left[ 6x - \frac{2x^3}{3} \right]_{-1}^{1} \) \( = \left( 6(1) - \frac{2(1)^3}{3} \right) - \left( 6(-1) - \frac{2(-1)^3}{3} \right) \) \( = \left( 6 - \frac{2}{3} \right) - \left( -6 + \frac{2}{3} \right) = 6 - \frac{2}{3} + 6 - \frac{2}{3} = 12 - \frac{4}{3} = \frac{36-4}{3} = \frac{32}{3} \) 3. \( \int_{1}^{3} (3-x)^2 dx = \left[ -\frac{(3-x)^3}{3} \right]_{1}^{3} = -\frac{(3-3)^3}{3} - \left( -\frac{(3-1)^3}{3} \right) = 0 - \left( -\frac{2^3}{3} \right) = \frac{8}{3} \) Finally, we sum these results and multiply by \( \frac{1}{16} \): \( A = \frac{1}{16} \left[ \frac{8}{3} + \frac{32}{3} + \frac{8}{3} \right] = \frac{1}{16} \left[ \frac{8+32+8}{3} \right] = \frac{1}{16} \left[ \frac{48}{3} \right] = \frac{1}{16} [16] = 1 \) Since the total area under the curve is 1, \( f(x) \) is indeed a valid probability density function.
In simple words: To check if this function is a proper probability density function, we need to make sure the total area under its graph equals 1. We do this by adding up the areas of three parts of the function using integration. After doing the math, we find the total area is indeed 1.

🎯 Exam Tip: When integrating piecewise functions for PDF verification, ensure the limits of integration for each piece are correct and that the sum of all definite integrals equals 1.

 

Question 8. A continuous random variable X has the following distribution function: Find (i) k and (ii) the probability density function.\[ F(x)= \begin{cases} 0 & \text{if } x \le 1 \\ k(x-1)^4 & \text{if } 1 < x < 3 \\ 1 & \text{if } x \ge 3 \end{cases} \]Answer:
(i) To find the value of \(k\), we use the properties of a cumulative distribution function (CDF). A key property is that \( F(x) \) must be continuous and must reach 1 at the upper limit of its range. At \( x=3 \), the value of \( F(x) \) should be 1. So, we set the middle part of the function equal to 1 when \( x=3 \): \( k(3-1)^4 = 1 \) \( k(2)^4 = 1 \) \( 16k = 1 \) Solving for \(k\): \( k = \frac{1}{16} \)
(ii) To find the probability density function (PDF), \( f(x) \), we differentiate the CDF, \( F(x) \), with respect to \( x \). \( f(x) = \frac{d}{dx} F(x) \) For the interval \( 1 < x < 3 \): \( f(x) = \frac{d}{dx} \left[ k(x-1)^4 \right] \) Substitute the value of \( k = \frac{1}{16} \): \( f(x) = \frac{d}{dx} \left[ \frac{1}{16}(x-1)^4 \right] \) Using the power rule and chain rule for differentiation: \( f(x) = \frac{1}{16} \cdot 4(x-1)^{4-1} \cdot \frac{d}{dx}(x-1) \) \( f(x) = \frac{4}{16}(x-1)^3 \cdot 1 \) \( f(x) = \frac{1}{4}(x-1)^3 \) For other values of \( x \), \( f(x) = 0 \). So, the probability density function is: \[ f(x)= \begin{cases} \frac{1}{4}(x-1)^3 & \text{for } 1 < x < 3 \\ 0 & \text{otherwise} \end{cases} \] This function describes the likelihood of \( X \) taking specific values within its range.
In simple words: (i) A distribution function must reach 1 at its highest point. We use this rule to find \(k\) by setting the function part equal to 1 at \(x=3\) and solving. (ii) To get the probability density function, we simply take the derivative (the slope) of the distribution function. This tells us how likely different values are.

🎯 Exam Tip: Remember that a CDF must be continuous and increase from 0 to 1. The PDF is the derivative of the CDF, and its integral over the entire range must be 1. These relationships are fundamental for these types of problems.

 

Question 9. The length of time (in minutes) that a certain person speaks on the telephone is found to be random phenomenon, with a probability function specified by the probability density function f(x) as\[ f(x)= \begin{cases} Ae^{-x/5} & \text{for } x \ge 0 \\ 0 & \text{otherwise} \end{cases} \]Answer:
(a) To find the value of \(A\) that makes \(f(x)\) a probability density function (PDF), we use the property that the total area under the PDF curve must be equal to 1. This means the integral of \(f(x)\) over its entire range must be 1. Since \(f(x)\) is defined for \(x \ge 0\), we integrate from 0 to infinity: \( \int_{-\infty}^{\infty} f(x)dx = 1 \) \( \int_{0}^{\infty} Ae^{-x/5} dx = 1 \) Factor out \(A\): \( A \int_{0}^{\infty} e^{-x/5} dx = 1 \) Evaluate the integral: \( A \left[ \frac{e^{-x/5}}{-1/5} \right]_{0}^{\infty} = 1 \) \( A \left[ -5e^{-x/5} \right]_{0}^{\infty} = 1 \) Substitute the limits: \( A \left( -5 \lim_{x \to \infty} e^{-x/5} - (-5e^{-0/5}) \right) = 1 \) Since \( \lim_{x \to \infty} e^{-x/5} = 0 \) and \( e^0 = 1 \): \( A ( -5(0) - (-5(1)) ) = 1 \) \( A (0 + 5) = 1 \) \( 5A = 1 \) Solve for \(A\): \( A = \frac{1}{5} \) This value ensures the total probability is 1.
(b) Now we calculate the probabilities using \( f(x) = \frac{1}{5}e^{-x/5} \) for \( x \ge 0 \). (i) Probability that the person talks more than 10 minutes, \( P(X > 10) \): \( P(X > 10) = \int_{10}^{\infty} f(x)dx = \int_{10}^{\infty} \frac{1}{5}e^{-x/5} dx \) \( = \frac{1}{5} \left[ -5e^{-x/5} \right]_{10}^{\infty} \) \( = - \left[ e^{-x/5} \right]_{10}^{\infty} \) \( = - (\lim_{x \to \infty} e^{-x/5} - e^{-10/5}) \) \( = - (0 - e^{-2}) = e^{-2} \) (ii) Probability that the person talks less than 5 minutes, \( P(X < 5) \): \( P(X < 5) = \int_{0}^{5} f(x)dx = \int_{0}^{5} \frac{1}{5}e^{-x/5} dx \) \( = \frac{1}{5} \left[ -5e^{-x/5} \right]_{0}^{5} \) \( = - \left[ e^{-x/5} \right]_{0}^{5} \) \( = - (e^{-5/5} - e^{-0/5}) \) \( = - (e^{-1} - 1) = 1 - e^{-1} \) (iii) Probability that the person talks between 5 and 10 minutes, \( P(5 < X < 10) \): \( P(5 < X < 10) = \int_{5}^{10} f(x)dx = \int_{5}^{10} \frac{1}{5}e^{-x/5} dx \) \( = \frac{1}{5} \left[ -5e^{-x/5} \right]_{5}^{10} \) \( = - \left[ e^{-x/5} \right]_{5}^{10} \) \( = - (e^{-10/5} - e^{-5/5}) \) \( = - (e^{-2} - e^{-1}) = e^{-1} - e^{-2} \) The exponential distribution is often used to model waiting times.
In simple words: (a) To find \(A\), we must make sure the total chance (area under the curve) for all possible talking times adds up to 1. We do this by integrating the function and solving for \(A\). (b) Once \(A\) is found, we use it to calculate the chance of talking (i) longer than 10 minutes, (ii) shorter than 5 minutes, and (iii) between 5 and 10 minutes, by integrating the function over those specific time ranges.

🎯 Exam Tip: When dealing with exponential probability distributions, remember that \( \int e^{ax} dx = \frac{1}{a} e^{ax} \). Also, carefully handle the negative signs during integration and evaluation of limits, especially for improper integrals going to infinity.

 

Question 10. Suppose that the time in minutes that a person has to wait at a certain station for a train is found to be a random phenomenon with a probability function specified by the distribution function\[ F(x)= \begin{cases} 0 & \text{for } x \le 0 \\ \frac{x}{2} & \text{for } 0 \le x < 1 \\ \frac{1}{2} & \text{for } 1 \le x < 2 \\ \frac{x}{4} & \text{for } 2 \le x < 4 \\ 1 & \text{for } x \ge 4 \end{cases} \]Answer:
(a) To check if the distribution function \( F(x) \) is continuous, we need to examine its behavior at the boundary points where its definition changes. - At \( x=0 \): \( \lim_{x \to 0^-} F(x) = 0 \), \( F(0) = \frac{0}{2} = 0 \), \( \lim_{x \to 0^+} F(x) = \frac{0}{2} = 0 \). It is continuous at \(x=0\). - At \( x=1 \): \( \lim_{x \to 1^-} F(x) = \frac{1}{2} \), \( F(1) = \frac{1}{2} \), \( \lim_{x \to 1^+} F(x) = \frac{1}{2} \). It is continuous at \(x=1\). - At \( x=2 \): \( \lim_{x \to 2^-} F(x) = \frac{1}{2} \), \( F(2) = \frac{2}{4} = \frac{1}{2} \), \( \lim_{x \to 2^+} F(x) = \frac{2}{4} = \frac{1}{2} \). It is continuous at \(x=2\). - At \( x=4 \): \( \lim_{x \to 4^-} F(x) = \frac{4}{4} = 1 \), \( F(4) = 1 \), \( \lim_{x \to 4^+} F(x) = 1 \). It is continuous at \(x=4\). Since \( F(x) \) is continuous at all its boundary points and is defined by continuous functions within each interval, the distribution function is indeed continuous on \( [0, 4] \). Now, we find the probability density function (PDF), \( f(x) \), by differentiating \( F(x) \) with respect to \( x \): - For \( 0 < x < 1 \): \( f(x) = \frac{d}{dx} \left( \frac{x}{2} \right) = \frac{1}{2} \) - For \( 1 < x < 2 \): \( f(x) = \frac{d}{dx} \left( \frac{1}{2} \right) = 0 \) - For \( 2 < x < 4 \): \( f(x) = \frac{d}{dx} \left( \frac{x}{4} \right) = \frac{1}{4} \) - For all other \( x \), \( f(x) = 0 \). So, the probability density function is: \[ f(x)= \begin{cases} \frac{1}{2} & \text{for } 0 < x < 1 \\ 0 & \text{for } 1 < x < 2 \\ \frac{1}{4} & \text{for } 2 < x < 4 \\ 0 & \text{otherwise} \end{cases} \]
(b) Now we calculate the probabilities using the PDF \( f(x) \): (i) Probability that a person will have to wait more than 3 minutes, \( P(X > 3) \): Since \( f(x) = \frac{1}{4} \) for \( 2 < x < 4 \): \( P(X > 3) = \int_{3}^{4} f(x)dx = \int_{3}^{4} \frac{1}{4} dx \) \( = \left[ \frac{1}{4}x \right]_{3}^{4} = \frac{1}{4}(4) - \frac{1}{4}(3) = 1 - \frac{3}{4} = \frac{1}{4} \) (ii) Probability that a person will have to wait less than 3 minutes, \( P(X < 3) \): This includes intervals \( (0,1) \), \( (1,2) \), and \( (2,3) \): \( P(X < 3) = \int_{0}^{1} f(x)dx + \int_{1}^{2} f(x)dx + \int_{2}^{3} f(x)dx \) \( = \int_{0}^{1} \frac{1}{2} dx + \int_{1}^{2} 0 dx + \int_{2}^{3} \frac{1}{4} dx \) \( = \left[ \frac{1}{2}x \right]_{0}^{1} + 0 + \left[ \frac{1}{4}x \right]_{2}^{3} \) \( = \frac{1}{2}(1-0) + \frac{1}{4}(3-2) = \frac{1}{2} + \frac{1}{4} = \frac{2}{4} + \frac{1}{4} = \frac{3}{4} \) (iii) Probability that a person will have to wait between 1 and 3 minutes, \( P(1 < X < 3) \): This includes intervals \( (1,2) \) and \( (2,3) \): \( P(1 < X < 3) = \int_{1}^{2} f(x)dx + \int_{2}^{3} f(x)dx \) \( = \int_{1}^{2} 0 dx + \int_{2}^{3} \frac{1}{4} dx \) \( = 0 + \left[ \frac{1}{4}x \right]_{2}^{3} = \frac{1}{4}(3-2) = \frac{1}{4} \) Understanding these probabilities helps in predicting waiting times at the station.
In simple words: (a) We first check if the waiting time function is smooth (continuous) by looking at its connection points. Then, we find its "density function" by taking the mathematical slope (derivative) of each part. (b) With the density function, we can calculate the chance of waiting (i) more than 3 minutes, (ii) less than 3 minutes, and (iii) between 1 and 3 minutes. We do this by calculating the area under the density curve for each time range.

🎯 Exam Tip: For piecewise CDFs, verify continuity at each breakpoint by checking if the left limit, right limit, and function value are all equal. For the PDF, differentiate each piece and ensure the ranges are correct.

 

Question 11.
Answer: The question content for Question 11 appears to be missing from the source material. Without the specific details or table for the random variable, a solution cannot be provided.

🎯 Exam Tip: When a question's necessary data (like a table or specific function) is missing, clearly state that the information is incomplete and a solution cannot be derived from the provided context.

 

Question 12. Explain What are the types of random variable.
Answer: Random variables are mainly divided into two types: discrete random variables and continuous random variables. 1. **Discrete Random Variables:** These variables can take a specific, countable number of values. These values are usually whole numbers that can be listed. For example, the number of heads when flipping a coin three times (0, 1, 2, or 3) is a discrete random variable. 2. **Continuous Random Variables:** These variables can take any value within a specific range. They can include decimals and fractions, and their values are typically found by measuring something. For example, a person's height or the amount of milk in a bottle are continuous random variables. Both types are very important for understanding and solving problems in mathematics and statistics.
In simple words: Random variables are either discrete or continuous. Discrete ones are for things you can count, like the number of red apples. Continuous ones are for things you measure, like how tall a person is or how long something takes.

🎯 Exam Tip: To distinguish, remember "discrete" means distinct, countable values, while "continuous" means any value within an interval, typically measured values.

 

Question 13. Define discrete random Variable
Answer: A discrete random variable is a variable whose possible values can be counted, either as a finite set of numbers or as an infinite but countable sequence. This means the values are separate and distinct, with gaps between them. For instance, the number of cars passing a point on a road in an hour is a discrete random variable, as it can only be 0, 1, 2, 3, and so on.
In simple words: A discrete random variable is a number that comes from a random event, and you can count its possible values. It's like counting whole items, not measuring something that can have tiny parts.

🎯 Exam Tip: Key phrases for discrete random variables are "countable values," "finite number of values," or "infinite sequence of countable real numbers."

 

Question 14. What do you understand by continuous random variable?
Answer: A continuous random variable is a variable that can take any value within a given interval or range on the real number line. Unlike discrete variables, its possible values cannot be counted individually. Instead, they are measured. For example, the exact temperature of a room, a person's weight, or the time it takes to complete a task are all continuous random variables, as they can have any value including fractions and decimals within a certain range.
In simple words: A continuous random variable is a number from a random event that can be any value within a certain range, not just specific numbers. Think of it like measuring something, where the value can be very precise.

🎯 Exam Tip: Remember that continuous variables are typically measured and can take on fractional or decimal values, while discrete variables are counted and usually take on integer values.

 

Question 15. Describe what is meant by a random variable
Answer: A random variable is a function that assigns a numerical value to each possible outcome of a random experiment or event. It is a way to turn non-numerical results (like "heads" or "tails") into numbers that we can use for mathematical analysis. For example, if we flip a coin, we could define a random variable \(X\) such that \(X=1\) if it's heads and \(X=0\) if it's tails. This allows us to work with probabilities and statistics more easily. It is also sometimes called a chance variable or stochastic variable.
In simple words: A random variable is simply a number that we assign to the result of a random event. It helps us use numbers to describe things that happen by chance, like how many times a certain outcome occurs.

🎯 Exam Tip: The core idea of a random variable is the assignment of numerical values to outcomes of a random experiment, allowing for mathematical analysis of probabilities.

 

Question 16. Distinguish between discrete and continuous random variable.
Answer: A random variable can be either discrete or continuous, depending on the type of values it can take. Here's how they are different:

In simple words: A discrete variable counts things (like number of cars), while a continuous variable measures things (like height or time). Discrete values are separate, but continuous values can be anywhere along a scale.

🎯 Exam Tip: When distinguishing between concepts, always provide clear, contrasting points for each term. Using examples helps illustrate the difference effectively.

 

Question 17. Explain the distribution function of random variable.
Answer: The distribution function of a random variable, often written as \( F_X(x) \) or simply \( F(x) \), tells us the probability that the random variable \( X \) will take a value less than or equal to a certain value \( x \). For a discrete random variable \( X \), its cumulative distribution function is defined by \( F_X(x) = P(X \le x) \), which means summing up all probabilities \( p(x_i) \) for values \( x_i \) that are less than or equal to \( x \). It helps to understand how probabilities build up as the variable's value increases.
In simple words: A distribution function shows the chance that a random number will be less than or equal to a specific value. It helps us see the overall pattern of the probabilities.

🎯 Exam Tip: Remember that the distribution function is cumulative, meaning it adds up probabilities. It always starts at 0 and goes up to 1, showing the total probability up to a certain point.

 

Question 18. Explain the terms (i) probability Mass function (ii) probability density function and (iii) probability distribution.
Answer:
(i) A probability mass function (PMF), denoted by \( P_X(x) \) or \( p(x) \), is used for discrete random variables. It gives the probability that a discrete random variable \( X \) is exactly equal to some value \( x_i \). This means \( P(X = x_i) = p_i \) or \( p(x) \). For all other values not in the set of possible outcomes, the probability is 0. The sum of all probabilities for a PMF must equal 1.
(ii) A probability density function (PDF), denoted by \( f_X(x) \) or \( f(x) \), is used for continuous random variables. It does not directly give probabilities for single points (which are 0 for continuous variables). Instead, the probability that a continuous random variable \( X \) falls within an interval \( [t_1, t_2] \) is found by integrating the PDF over that interval: \( P(t_1 \le X \le t_2) = \int_{t_1}^{t_2} f_x(x)dx \). The PDF helps describe the likelihood of values falling within ranges.
(iii) A probability distribution describes all the possible values a random variable can take and how often each value occurs, or the probability of each range of values. For a random variable \( X \) with distinct values \( X_1, X_2, \ldots, X_n \) and their corresponding probabilities \( p_1, p_2, p_3, \ldots, p_n \), the sum of all these probabilities must be equal to 1, i.e., \( \sum_{i=1}^{n} P_i = 1 \). This summarizes the entire probabilistic behavior of the variable.
In simple words: PMF is for counting individual chances, PDF is for finding chances over a range for continuous numbers, and probability distribution is the full picture of all possible outcomes and their likelihoods.

🎯 Exam Tip: Remember that PMF gives exact probabilities for discrete values, while PDF gives probabilities for intervals of continuous values through integration. The sum or integral of all probabilities must always be 1.

 

Question 19. What are the properties of (i) a probability mass function (PMF) and (ii) a probability density function (PDF)?
Answer:
(i) For a probability mass function (PMF), \( p(x) \), these conditions must be true:
1. The probability for each value must be non-negative: \( p(x) \ge 0 \) for all \( x_i \).
2. The sum of all probabilities for all possible values must be exactly 1: \( \sum_{i=1}^{\infty} p(x_i) = 1 \). This ensures that one of the outcomes must occur.
(ii) For a probability density function (PDF), \( f(x) \), these conditions must be true:
1. The function's value must be non-negative for all \( x \): \( f(x) \ge 0 \) for all \( x \).
2. The total area under the curve of the function must be exactly 1 when integrated over its entire range: \( \int_{-\infty}^{\infty} f(x) dx = 1 \). This represents the total probability of all possible outcomes.
In simple words: For both types of probability functions, all probabilities must be zero or positive. Also, the total chance of all possible events happening must always add up to 1.

🎯 Exam Tip: These properties are fundamental to probability theory. Always check that the conditions for non-negativity and the sum/integral to 1 are met when defining or working with PMFs and PDFs.

 

Question 20. State the Properties of distribution function.
Answer: The distribution function, \( F_X(x) \) or \( F(x) \), has several key properties:
(i) The value of the function always stays between 0 and 1, inclusive: \( 0 \le F(x) \le 1 \) for all \( -\infty < x < \infty \).
(ii) As \( x \) approaches negative infinity, \( F(x) \) approaches 0 (no probability has accumulated yet): \( F(-\infty) = \lim_{x \rightarrow -\infty} F(x) = 0 \). As \( x \) approaches positive infinity, \( F(x) \) approaches 1 (all probability has accumulated): \( F(+\infty) = \lim_{x \rightarrow \infty} F(x) = 1 \).
(iii) \( F(x) \) is a monotone, non-decreasing function. This means if \( a < b \), then \( F(a) \le F(b) \). The probability never goes down as \( x \) increases.
(iv) \( F(x) \) is continuous from the right. This means that as \( h \) approaches 0 from the positive side, \( F(x + h) \) approaches \( F(x) \): \( \lim_{h \rightarrow 0} F(x + h) = F(x) \).
(v) The derivative of \( F(x) \) with respect to \( x \) gives the probability density function \( f(x) \), and this must be non-negative: \( F'(x) = \frac{d}{dx} F(x) = f(x) \ge 0 \).
(vi) If \( F'(x) = \frac{d}{dx} F(x) = f(x) \), then \( dF(x) = f(x)dx \). This \( dF(x) \) is known as the probability differential of \( X \).
(vii) The probability that \( X \) falls within an interval \( [a, b] \) can be found using the distribution function:
\( P(a \le x \le b) = \int_{a}^{b} f(x)dx \)
\( \implies P(a \le x \le b) = \int_{-\infty}^{b} f(x)dx – \int_{-\infty}^{a} f(x)dx \)
\( \implies P(a \le x \le b) = P(X \le b) – P(X \le a) \)
\( \implies P(a \le x \le b) = F(b) – F(a) \). This shows how the function helps find probabilities for intervals.
In simple words: A distribution function always ranges from 0 to 1. It only increases or stays the same as the value of \( x \) goes up. It helps us find the chance of a random number falling within any given range.

🎯 Exam Tip: Understanding these properties is crucial for working with probability distributions. Especially remember that \( F(x) \) is non-decreasing and spans the entire probability range from 0 to 1.