Samacheer Kalvi Class 12 Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 1.3

 

Question 1. Find the expected value for the random variable of an unbiased die
Answer: When an unbiased die is rolled, each number from 1 to 6 has an equal chance of appearing. This means the probability for each number (1, 2, 3, 4, 5, or 6) is \( \frac{1}{6} \). To find the expected value, we multiply each possible outcome by its probability and then add all these results together. This gives us the average outcome we would expect over many rolls.

The expected value E(x) is calculated as: \( E(x) = \sum_{i=1}^{n} p_i x_i \)
\( = \left( \frac{1}{6} \times 1 \right) + \left( \frac{1}{6} \times 2 \right) + \left( \frac{1}{6} \times 3 \right) + \left( \frac{1}{6} \times 4 \right) + \left( \frac{1}{6} \times 5 \right) + \left( \frac{1}{6} \times 6 \right) \)
\( = \frac{1}{6} [1+2+3+4+5+6] \)
\( = \frac{1}{6} [21] \)
\( = \frac{21}{6} \)
\( = \frac{7}{2} \)
\( = 3.5 \) Therefore, the expected value of the number on a die is 3.5.In simple words: When you roll a fair die, the average number you expect to get over many rolls is 3.5. We find this by adding up each possible number multiplied by how likely it is to appear.

🎯 Exam Tip: Remember that the expected value doesn't have to be one of the actual outcomes. It's an average, so a die can't roll a 3.5, but it's the theoretical average over many trials.

 

Question 2. Let X be a random variable defining number of students getting A grade. Find the expected value of X from the given table
Answer: Here, X is a random variable representing the number of students who got an A grade, and we are given a table with the probabilities for each number of students. To find the expected value, we multiply each possible number of students by its probability and then sum these products. This will give us the average number of students expected to get an A grade.

The random variable X takes values 0, 1, 2, 3. Expected value \( E(x) = \sum p_i x_i \)
\( = (0.2 \times 0) + (0.1 \times 1) + (0.4 \times 2) + (0.3 \times 3) \)
\( = 0 + 0.1 + 0.8 + 0.9 \)
\( = 1.8 \) Therefore, the expected value is 1.8.In simple words: Based on the given chances, we expect an average of 1.8 students to get an A grade. Even though you can't have 1.8 students, this is the calculated average.

🎯 Exam Tip: Always double-check that the probabilities in the table add up to 1, which is a key property of a probability mass function.

 

Question 3. The following table is describing about the probability mass function of the random variable X. Find the standard deviation of x.
Answer: To find the standard deviation, we first need to calculate the expected value (mean) and then the variance. The standard deviation tells us how spread out the values of the random variable are from the mean. A larger standard deviation means the values are more spread out.

Let X be the random variable taking values 3, 4, 5. First, calculate the expected value \( E(x) \): \( E(x) = \sum p_i x_i \)
\( = (0.2 \times 3) + (0.3 \times 4) + (0.5 \times 5) \)
\( = 0.6 + 1.2 + 2.5 \)
\( = 4.3 \) Next, calculate the expected value of \( x^2 \), denoted as \( E(x^2) \): \( E(x^2) = \sum p_i x_i^2 \)
\( = (0.2 \times 3^2) + (0.3 \times 4^2) + (0.5 \times 5^2) \)
\( = (0.2 \times 9) + (0.3 \times 16) + (0.5 \times 25) \)
\( = 1.8 + 4.8 + 12.5 \)
\( = 19.1 \) Now, calculate the variance \( Var(x) \): \( Var(x) = E(x^2) - [E(x)]^2 \)
\( = 19.1 - (4.3)^2 \)
\( = 19.1 - 18.49 \)
\( = 0.61 \) Finally, calculate the standard deviation (S.D.): \( \text{S.D.} = \sqrt{ Var(x) } \)
\( = \sqrt{ 0.61 } \)
\( \sigma \approx 0.781 \) Thus, the standard deviation is approximately 0.781.In simple words: We first found the average value (mean) of X, which is 4.3. Then we calculated how much the values spread out from this average by finding the variance, which was 0.61. Taking the square root of the variance gives us the standard deviation, about 0.781, showing how typical values differ from the mean.

🎯 Exam Tip: Remember the formula for variance: \( Var(x) = E(x^2) - [E(x)]^2 \). This is often easier than \( \sum p_i (x_i - E(x))^2 \), especially with complex numbers.

 

Question 4. Let X be a continuous random variable with probability density function \( f_x (x) = \left\{\begin{array}{l} 2 x, 0 \leq x \leq 1 \\ 0, \text { otherwise } \end{array}\right. \)
Answer: For a continuous random variable, the expected value (mean) is found by integrating \( x \cdot f_x(x) \) over all possible values of x. Since the function is only non-zero between 0 and 1, we only need to integrate over that range. This integral helps us find the weighted average of all possible values X can take. The expected value E(x) is given by: \( E(x) = \int_{-\infty}^{\infty} x f(x) dx \) Here, \( f(x) = 2x \) for \( 0 \le x \le 1 \) and \( 0 \) otherwise. So, we calculate E(x) as:
\( E(x) = \int_{0}^{1} x(2x) dx \)
\( = \int_{0}^{1} 2x^2 dx \)
\( = 2 \left[ \frac{x^3}{3} \right]_{0}^{1} \)
\( = 2 \left( \frac{1^3}{3} - \frac{0^3}{3} \right) \)
\( = 2 \left( \frac{1}{3} - 0 \right) \)
\( = \frac{2}{3} \) Therefore, the expected value \( E(x) = \frac{2}{3} \).In simple words: For a continuous random variable, the expected value is found by doing a special kind of sum called an integral. For this specific function, the average value we expect is \( \frac{2}{3} \).

🎯 Exam Tip: Remember that for a continuous probability density function, the expected value is calculated using an integral, not a summation.

 

Question 5. Let X be a continuous random variable with probability density function \( f_x (x) = \left\{\begin{array}{l} 2 x, 0 \leq x \leq 1 \\ 0, \text { otherwise } \end{array}\right. \) Find the mean and variance of X.
Answer: To find the mean (expected value) and variance of a continuous random variable, we use integration. The mean tells us the average value, while the variance measures how spread out the values are from the mean. We will use the probability density function \( f(x) = 3x^{-4} \) for \( x \ge 1 \) and \( 0 \) otherwise, as implied by the given solution steps. First, let's find the mean \( E(x) \): \( E(x) = \int_{-\infty}^{\infty} x f(x) dx \)
\( = \int_{1}^{\infty} x (3x^{-4}) dx \)
\( = \int_{1}^{\infty} 3x^{-3} dx \)
\( = 3 \left[ \frac{x^{-3+1}}{-3+1} \right]_{1}^{\infty} \)
\( = 3 \left[ \frac{x^{-2}}{-2} \right]_{1}^{\infty} \)
\( = -\frac{3}{2} \left[ \frac{1}{x^2} \right]_{1}^{\infty} \)
\( = -\frac{3}{2} \left( \lim_{x \to \infty} \frac{1}{x^2} - \frac{1}{1^2} \right) \)
\( = -\frac{3}{2} (0 - 1) \)
\( = \frac{3}{2} \) So, the mean \( E(x) = \frac{3}{2} \). Next, let's find \( E(x^2) \) to calculate the variance: \( E(x^2) = \int_{-\infty}^{\infty} x^2 f(x) dx \)
\( = \int_{1}^{\infty} x^2 (3x^{-4}) dx \)
\( = \int_{1}^{\infty} 3x^{-2} dx \)
\( = 3 \left[ \frac{x^{-2+1}}{-2+1} \right]_{1}^{\infty} \)
\( = 3 \left[ \frac{x^{-1}}{-1} \right]_{1}^{\infty} \)
\( = -3 \left[ \frac{1}{x} \right]_{1}^{\infty} \)
\( = -3 \left( \lim_{x \to \infty} \frac{1}{x} - \frac{1}{1} \right) \)
\( = -3 (0 - 1) \)
\( = 3 \) So, \( E(x^2) = 3 \). Finally, calculate the variance \( Var(x) \): \( Var(x) = E(x^2) - [E(x)]^2 \)
\( = 3 - \left( \frac{3}{2} \right)^2 \)
\( = 3 - \frac{9}{4} \)
\( = \frac{12}{4} - \frac{9}{4} \)
\( = \frac{3}{4} \) Therefore, the variance \( Var(x) = \frac{3}{4} \).In simple words: We calculated the average value (mean) of X, which is \( \frac{3}{2} \). Then, to find how spread out the values are, we used another integral to get \( E(x^2) \) and subtracted the square of the mean. This gave us a variance of \( \frac{3}{4} \), showing the spread of the data.

🎯 Exam Tip: Pay close attention to the limits of integration for continuous random variables, as they define the region where the probability density function is non-zero.

 

Question 6. In an investment, a man can make a profit of Rs 5,000 with a probability of 0.62 or a loss of Rs 8,000 with a probability of 0.38. Find the expected gain.
Answer: To find the expected gain, we multiply each possible outcome (profit or loss) by its probability and then add these values together. A loss is represented by a negative value. The expected gain tells us the average profit or loss one might expect over many similar investments.

Let X be the random variable representing the gain in an investment. Expected gain \( E(X) = \sum p_i x_i \)
\( = (0.62 \times 5000) + (0.38 \times -8000) \)
\( = 3100 + (-3040) \)
\( = 3100 - 3040 \)
\( = 60 \) Therefore, the expected gain is Rs 60.In simple words: We calculated the average money someone expects to earn or lose from this investment. Even though there's a big profit or a big loss, on average, after considering the chances, the person expects to gain Rs 60.

🎯 Exam Tip: Always represent a loss as a negative value in your calculations for expected gain or profit, to ensure the correct weighted average.

 

Question 7. What are the properties of mathematical expectation?
Answer: Mathematical expectation, or the expected value, has several important properties that help us work with random variables and constants. These rules make it easier to calculate expected values for different situations.
(i) If 'a' is a constant, then the expected value of 'a' is 'a' itself. For example, \( E(5) = 5 \).
(ii) If 'a' is a constant and 'X' is a random variable, then the expected value of 'aX' is 'a' times the expected value of 'X'. For example, \( E(2X) = 2E(X) \).
(iii) If 'a' and 'b' are constants and 'X' is a random variable, then the expected value of \( aX + b \) is 'a' times \( E(X) \) plus 'b'. For example, \( E(2X + 3) = 2E(X) + 3 \).
(iv) If 'X' is a non-negative random variable (meaning \( X \ge 0 \)), then its expected value \( E(X) \) will also be non-negative (i.e., \( E(X) \ge 0 \)).
(v) The variance of a constant 'a' is 0, i.e., \( V(a) = 0 \). This is because a constant does not change, so it has no spread or variation.
(vi) If 'X' is a random variable and 'a' and 'b' are constants, then the variance of \( aX + b \) is \( a^2 \) times the variance of 'X'. The constant 'b' does not affect the variance. So, \( V(aX + b) = a^2 V(X) \).In simple words: These rules tell us how the average value (expectation) changes when we add, subtract, multiply, or divide a random variable by a fixed number. For example, the average of a fixed number is just that number, and multiplying a random variable by a constant multiplies its average by the same constant.

🎯 Exam Tip: Clearly distinguish between properties of expectation \( E(X) \) and properties of variance \( V(X) \), as constants affect them differently (e.g., \( E(X+b) = E(X)+b \) but \( V(X+b) = V(X) \)).

 

Question 8. What do you understand by mathematical expectation?
Answer: Mathematical expectation, often called the expected value or mean, is the average value of a random variable. It tells us what outcome we can expect on average if an experiment is repeated many times. For example, if you play a game, the expected value tells you how much money you would win or lose per game over a very long period. It's a weighted average where each possible outcome is multiplied by its probability.In simple words: Mathematical expectation is simply the average value you would expect from something that happens randomly, like the average score on a dice roll over many tries.

🎯 Exam Tip: Define mathematical expectation as a "weighted average" of possible outcomes, as this highlights the role of probability in its calculation.

 

Question 9. How do you define variance in terms of Mathematical expectation?
Answer: Variance is a measure of how spread out the values of a random variable are from its mean (mathematical expectation). It can be defined using mathematical expectation as the expected value of the squared difference between the random variable and its mean. In simpler terms, it calculates the average of how far each value is from the mean, but squared to make all differences positive and give more weight to larger deviations. The formula for variance \( Var(X) \) in terms of mathematical expectation is: \( Var(X) = E[(X - E(X))^2] \) This formula shows that variance is the expected value of the squared deviations from the mean. A more common and computationally easier formula derived from this is: \( Var(X) = E(X^2) - [E(X)]^2 \) Where \( E(X^2) \) is the expected value of \( X \) squared, and \( [E(X)]^2 \) is the square of the expected value of \( X \).In simple words: Variance tells you how much the numbers in a set usually differ from their average. We find it by taking the average of the squared distances from each number to the average of all numbers.

🎯 Exam Tip: Always use the formula \( Var(X) = E(X^2) - [E(X)]^2 \) for calculations, as it simplifies the process and reduces errors, especially when dealing with discrete data.

 

Question 10. Define mathematical expectation in terms of discrete random variable?
Answer: For a discrete random variable, mathematical expectation is defined as the sum of the products of each possible value of the variable and its corresponding probability. Imagine all the possible outcomes you could get, and for each outcome, you multiply it by how likely it is to happen. Then, you add all these results together. This sum gives you the average value you'd expect to see over many trials. If X is a discrete random variable with a probability mass function \( p(x) \) (meaning \( p(x) \) is the probability that X takes a specific value x), then its expected value is defined by: \( E(X) = \sum_x x p(x) \) Here, the sum is taken over all possible values 'x' that the random variable X can take. This formula helps to calculate the mean or average outcome of a discrete random process.In simple words: For a variable that takes on separate, clear values, the expected value is found by multiplying each value by its chance of happening, and then adding all those results up.

🎯 Exam Tip: Clearly state that for discrete variables, the expectation is a "summation" of \( x \cdot p(x) \), contrasting it with continuous variables that use integration.

 

Question 11. State the definition of mathematical expectation using continuous random variable.
Answer: For a continuous random variable, mathematical expectation (mean) is defined using an integral. Instead of summing individual products like for discrete variables, we integrate the product of the variable's value and its probability density function over all possible values it can take. This integral represents the weighted average of all outcomes, where the density function gives the relative likelihood of each outcome. If X is a continuous random variable and \( f_x(x) \) is its probability density function, then the expected value \( E(X) \) is defined by: \( E(X) = \int_{-\infty}^{\infty} x f(x) dx \) This formula calculates the "long-run average" or the mean value that X would take if we observed it an infinite number of times. The integral is crucial for continuous variables because they can take any value within a range, unlike discrete variables that have distinct, separate values.In simple words: For a variable that can take any value within a range (continuous), the expected value is found by using a special "sum" called an integral, where we multiply each value by its probability density.

🎯 Exam Tip: Highlight that the core difference between discrete and continuous expectation is the use of "summation" versus "integration," respectively.

 

Question 12. In a business venture a man can make a profit of Rs 2,000 with a probability of 0.4 or have a loss of Rs 1,000 with a probability of 0.6. What is his expected, variance and standard deviation of profit?
Answer: We need to calculate the expected profit, variance, and standard deviation for this business venture. The expected profit tells us the average outcome, while the variance and standard deviation measure how much the actual profit might vary from this average. Remember to treat losses as negative values in your calculations.

Let X be the random variable representing the profit in the business. First, calculate the Expected Profit \( E(X) \): \( E(X) = \sum x_i p_i(x) \)
\( = (0.4 \times 2000) + (0.6 \times -1000) \)
\( = 800 - 600 \)
\( = 200 \) So, the expected value of profit is Rs 200. Next, calculate \( E(X^2) \): \( E(X^2) = \sum x_i^2 p_i(x) \)
\( = [(2000)^2 \times 0.4] + [(-1000)^2 \times 0.6] \)
\( = (4000000 \times 0.4) + (1000000 \times 0.6) \)
\( = 1600000 + 600000 \)
\( = 2200000 \) Now, calculate the Variance \( Var(X) \): \( Var(X) = E(X^2) - [E(X)]^2 \)
\( = 2200000 - (200)^2 \)
\( = 2200000 - 40000 \)
\( = 2160000 \) So, the variance of his profit is Rs 21,60,000. Finally, calculate the Standard Deviation (S.D.): \( \sigma = \sqrt{ Var(X) } \)
\( = \sqrt{ 2160000 } \)
\( \sigma \approx 1469.69 \) Thus, the standard deviation of his profit is approximately Rs 1,469.69.In simple words: On average, the man expects to make a profit of Rs 200. However, the profit can vary quite a lot from this average. The variance (Rs 2,160,000) and standard deviation (Rs 1,469.69) show how much spread there is in his possible profits or losses.

🎯 Exam Tip: When calculating variance, ensure you square the *individual outcomes* \( (X^2) \) before multiplying by probability for \( E(X^2) \), and then square the *overall expected value* \( (E(X))^2 \) for the final step.

 

Question 13. The number of miles an automobile tire lasts before it reaches a critical point in tread wear can be represented by a p.d.f. \( f(x)=\left\{\begin{array}{l} \frac{1}{30} e^{-\frac{x}{30}}, \text { for } x>0 \\ 0 \text { for } x \leq 0 \end{array}\right. \) Find the expected number of miles (in thousands) a tire would last until it point.
Answer: To find the expected number of miles a tire would last, we need to calculate the expected value \( E(x) \) for this continuous random variable. This involves integrating the product of \( x \) and the given probability density function \( f(x) \) over the valid range of \( x \). This expected value gives us the average lifespan of the tire in thousands of miles. The expected value \( E(x) \) is given by: \( E(x) = \int_{-\infty}^{\infty} xf(x) dx \) Given \( f(x) = \frac{1}{30} e^{-\frac{x}{30}} \) for \( x > 0 \), and \( 0 \) otherwise.
\( E(x) = \int_{0}^{\infty} x \left( \frac{1}{30} e^{-\frac{x}{30}} \right) dx \)
\( = \frac{1}{30} \int_{0}^{\infty} x e^{-\frac{x}{30}} dx \) We can use the Gamma Integral formula: \( \int_{0}^{\infty} x^n e^{-ax} dx = \frac{n!}{a^{n+1}} \) In our case, \( n=1 \) and \( a = \frac{1}{30} \).
\( E(x) = \frac{1}{30} \times \frac{1!}{\left(\frac{1}{30}\right)^{1+1}} \)
\( = \frac{1}{30} \times \frac{1}{\left(\frac{1}{30}\right)^2} \)
\( = \frac{1}{30} \times \frac{1}{\frac{1}{900}} \)
\( = \frac{1}{30} \times 900 \)
\( = 30 \) Therefore, the expected number of miles a tire would last is 30 thousand miles.In simple words: Using a special math trick called the Gamma Integral, we found that on average, these tires are expected to last for 30,000 miles.

🎯 Exam Tip: Recognizing standard integral forms like the Gamma Integral can save significant time on calculus-based probability questions.

 

Question 14. A person tosses a coin and is to receive Rs 4 for a head and is to pay Rs 2 for a tail. Find the expectation and variance of his gains.
Answer: This problem involves a simple coin toss with specific payouts for heads and tails. We need to find the expected gain (average gain over many tosses) and the variance (how much the actual gain might vary from the expected gain). Since a coin is tossed, the probability of getting a head is \( \frac{1}{2} \) and the probability of getting a tail is also \( \frac{1}{2} \). Let X denote the amount the person receives (gain) in the game. If a head occurs, X = Rs 4. If a tail occurs, X = -Rs 2 (since it's a payment/loss). The probability distribution is: \( P(X = 4) = P(\text{of getting a head}) = \frac{1}{2} \) \( P(X = -2) = P(\text{of getting a tail}) = \frac{1}{2} \) First, calculate the Expected Gain \( E(X) \): \( E(X) = \sum x_i p_i(x) \)
\( = \left( 4 \times \frac{1}{2} \right) + \left( -2 \times \frac{1}{2} \right) \)
\( = 2 + (-1) \)
\( = 1 \) So, the expected gain is Rs 1. Next, calculate \( E(X^2) \): \( E(X^2) = \sum x_i^2 p_i(x) \)
\( = \left[ (4)^2 \times \frac{1}{2} \right] + \left[ (-2)^2 \times \frac{1}{2} \right] \)
\( = \left[ 16 \times \frac{1}{2} \right] + \left[ 4 \times \frac{1}{2} \right] \)
\( = 8 + 2 \)
\( = 10 \) Now, calculate the Variance \( Var(X) \): \( Var(X) = E(X^2) - [E(X)]^2 \)
\( = 10 - (1)^2 \)
\( = 10 - 1 \)
\( = 9 \) Therefore, the variance of his gain is 9.In simple words: On average, this person expects to gain Rs 1 per coin toss. The variance of 9 shows that the actual money gained or lost can vary quite a bit from this average.

🎯 Exam Tip: Always correctly assign negative values to losses when setting up the random variable for financial outcomes. This ensures accurate calculation of expected gain and variance.

 

Question 15. Let X be a random variable and Y = 2X + 1. What is the variance of Y if variance of X is 5?
Answer: We are given the variance of a random variable X and a new random variable Y that is defined as a linear transformation of X. We need to find the variance of Y using the properties of variance. A key property states that if \( Y = aX + b \), then \( Var(Y) = a^2 Var(X) \). The constant 'b' (in this case, +1) does not affect the variance because it just shifts all the values by the same amount, without changing how spread out they are. Given: Variance of X, \( Var(X) = 5 \) The relationship between Y and X is \( Y = 2X + 1 \). Using the property of variance for a linear transformation, \( Var(aX + b) = a^2 Var(X) \): Here, \( a = 2 \) and \( b = 1 \).
\( Var(Y) = Var(2X + 1) \)
\( = (2)^2 Var(X) \)
\( = 4 \times Var(X) \) Substitute the given value of \( Var(X) \):
\( Var(Y) = 4 \times 5 \)
\( = 20 \) Therefore, the variance of Y is 20.In simple words: If you multiply a random variable by a number (like 2) and then add another number (like 1), only the multiplication changes how spread out the data is. The variance of X was 5, so multiplying X by 2 made the variance of Y four times bigger, resulting in 20.

🎯 Exam Tip: Remember that adding a constant (like '+1' here) does not change the variance of a random variable, as it only shifts the distribution without affecting its spread. Only multiplication by a constant (which is squared) influences the variance.