Samacheer Kalvi Class 12 Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 1.1

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Miscellaneous Problems

 

Question 1. The probability function of a random variable X is given by\[ p(x)= \begin{cases} \frac{1}{4}, & \text{for } x=-2 \\ \frac{1}{4}, & \text{for } x=0 \\ \frac{1}{2}, & \text{for } x=10 \\ 0, & \text{elsewhere} \end{cases} \]Evaluate the following probabilities.
(i) P(X ≤ 0)
(ii) P(X ≤ 0)
(iii) P(|X| ≤ 2)
(iv) P(0 ≤ X ≤ 10)
Answer:

(i) \(P(X \le 0) = P(X = -2) + P(X = 0)\)
\( = \frac{1}{4} + \frac{1}{4} \)
\( = \frac{2}{4} \)
\( = \frac{1}{2} \)
(ii) \(P(X < 0) = P(X = -2)\)
\( = \frac{1}{4} \)
(iii) \(P(|X| \le 2) = P(-2 \le X \le 2)\)
\( = P(X = -2) + P(X = 0)\)
\( = \frac{1}{4} + \frac{1}{4} \)
\( = \frac{2}{4} \)
\( = \frac{1}{2} \)
(iv) \(P(0 \le X \le 10) = P(X = 0) + P(X = 10)\)
\( = \frac{1}{4} + \frac{1}{2} \)
\( = \frac{1+2}{4} \)
\( = \frac{3}{4} \)
In simple words: We find the total probability by adding up the probabilities for each specific value of X. For (i) and (ii), even though the question repeats, the solution distinctively calculates for P(X ≤ 0) and P(X < 0), which covers probabilities up to and strictly less than zero.

🎯 Exam Tip: Always sum the individual probabilities for the given range of a discrete random variable. Remember that \( P(X \le a) \) includes \( X=a \), while \( P(X < a) \) does not.

 

Question 2. The probability function of a random variable X is given by\[ F(x)= \begin{cases} 0, & \text{if } x<0 \\ \frac{x}{8}, & \text{if } 0 \le x < 1 \\ \frac{1}{4} + \frac{x}{8}, & \text{if } 1 \le x < 2 \\ \frac{3}{4} + \frac{x}{12}, & \text{if } 2 \le x < 3 \\ 1, & \text{for } x \ge 3 \end{cases} \](a) Compute: (i) P(1 ≤ X ≤ 2) and (ii) P(X = 3).
(b) Is X a discrete random variable? Justify your answer.
Answer:The probability density function \( f(x) \) is found by differentiating the cumulative distribution function \( F(x) \). \[ f(x) = \frac{d[F(x)]}{dx} = \begin{cases} 0, & \text{if } x<0 \\ \frac{1}{8}, & \text{if } 0 \le x < 1 \\ \frac{1}{8}, & \text{if } 1 \le x < 2 \\ \frac{1}{12}, & \text{if } 2 \le x < 3 \\ 0, & \text{elsewhere} \end{cases} \] (a) (i) To find \( P(1 \le X \le 2) \), we can use the cumulative distribution function: \( P(1 \le X \le 2) = F(2) - F(1) \) We calculate \( F(2) \) using the third case \( \frac{3}{4} + \frac{x}{12} \) for \( x=2 \): \( F(2) = \frac{3}{4} + \frac{2}{12} = \frac{3}{4} + \frac{1}{6} = \frac{9+2}{12} = \frac{11}{12} \) We calculate \( F(1) \) using the second case \( \frac{1}{4} + \frac{x}{8} \) for \( x=1 \): \( F(1) = \frac{1}{4} + \frac{1}{8} = \frac{2+1}{8} = \frac{3}{8} \) Now, substitute these values back: \( P(1 \le X \le 2) = F(2) - F(1) = \frac{11}{12} - \frac{3}{8} \) To subtract these, find a common denominator, which is 24: \( P(1 \le X \le 2) = \frac{11 \times 2}{12 \times 2} - \frac{3 \times 3}{8 \times 3} = \frac{22}{24} - \frac{9}{24} = \frac{13}{24} \)
(a) (ii) \( P(X = 3) = 0 \) because the probability density function \( f(x) \) is not a step function; it's a continuous distribution where the probability at a single point is zero.
(b) X is not a discrete random variable. It is a continuous random variable because its cumulative distribution function \( F(x) \) is continuous, and its probability density function \( f(x) \) is defined by a continuous function, not by probabilities at distinct points. We can see its values vary continuously over certain ranges.
In simple words: For part (a), we use the given \( F(x) \) formula to find probabilities over a range by subtracting the \( F \) values at the start and end of the range. For a single point like \( X=3 \), the probability is zero for a continuous variable. For part (b), \( X \) is continuous because its values can smoothly change, unlike discrete variables which can only take specific, separate values.

🎯 Exam Tip: Remember that for a continuous random variable, the probability at any single exact point is always zero. Probabilities are only defined over intervals.

 

Question 3. The p.d.f. of X is defined as\[ f(x)= \begin{cases} k, & \text{for } 0Answer:Let X be a random variable, and for a probability density function, the total probability over its range must be 1. We know that \( \int_{-\infty}^{\infty} f(x)dx = 1 \) Here, the function is non-zero only between 0 and 4. So, \( \int_{0}^{4} f(x)dx = 1 \)
\( \implies \int_{0}^{4} k \, dx = 1 \)
\( \implies k[x]_{0}^{4} = 1 \)
\( \implies k[4-0] = 1 \)
\( \implies 4k = 1 \)
\( \implies k = \frac{1}{4} \) Now we need to calculate \( P(2 \le X \le 4) \): \( P(2 \le X \le 4) = \int_{2}^{4} f(x)dx \) Since \( f(x) = k = \frac{1}{4} \) for the range \( 0 < x \le 4 \), we use this value: \( P(2 \le X \le 4) = \int_{2}^{4} \frac{1}{4} \, dx \)
\( \implies P(2 \le X \le 4) = \frac{1}{4}[x]_{2}^{4} \)
\( \implies P(2 \le X \le 4) = \frac{1}{4}[4-2] \)
\( \implies P(2 \le X \le 4) = \frac{1}{4}[2] \)
\( \implies P(2 \le X \le 4) = \frac{1}{2} \)
In simple words: First, we find the value of 'k' by making sure that the total probability of all possible outcomes equals 1. Then, we use this 'k' value to find the probability within a specific range, just by integrating the probability density function over that range. This tells us the chance of X falling into that interval.

🎯 Exam Tip: For probability density functions, always ensure that the total area under the curve is 1, as this is how you typically find unknown constants like 'k'.

 

Question 4. The probability distribution function of a discrete random variable X is\[ f(x)= \begin{cases} 2k, & x=1 \\ 3k, & x=3 \\ 4k, & x=5 \\ 0, & \text{otherwise} \end{cases} \]where k is some constant. Find (a) k and (b) P(X > 2).
Answer:(a) For a probability distribution function of a discrete random variable, the sum of all probabilities must be 1. So, \( \sum P(x) = 1 \) \( P(X=1) + P(X=3) + P(X=5) = 1 \) \( 2k + 3k + 4k = 1 \) \( 9k = 1 \)
\( \implies k = \frac{1}{9} \)
(b) To find \( P(X > 2) \), we need to sum the probabilities for values of X greater than 2. From the given function, these values are \( X=3 \) and \( X=5 \). \( P(X > 2) = P(X=3) + P(X=5) \) \( = 3k + 4k \) \( = 7k \) Now, substitute the value of \( k = \frac{1}{9} \) that we found: \( P(X > 2) = 7 \left( \frac{1}{9} \right) = \frac{7}{9} \)
In simple words: For part (a), we find 'k' by adding up all the given probabilities and setting the sum equal to 1, because that's a basic rule for all probabilities. For part (b), we just add the probabilities for the values of X that are larger than 2, using the 'k' we just found.

🎯 Exam Tip: Always remember that the sum of probabilities for all possible outcomes in a discrete probability distribution must equal one. This is key for finding unknown constants.

 

Question 5. The probability distribution function of a discrete random variable X is\[ f(x)= \begin{cases} a+bx^{2}, & 0 \le x \le 1 \\ 0, & \text{otherwise} \end{cases} \]where a and b are some constants. Find (i) a and b if E(X) = 3/5 (ii) Var(X).
Answer:Let X be a continuous random variable with a density function. We know that for a probability density function, the total integral over its range must be 1. \( \int_{-\infty}^{\infty} f(x)dx = 1 \) Since \( f(x) \) is non-zero only between 0 and 1: \( \int_{0}^{1} (a+bx^{2})dx = 1 \)
\( \implies \left[ ax + \frac{bx^{3}}{3} \right]_{0}^{1} = 1 \)
\( \implies \left( a(1) + \frac{b(1)^{3}}{3} \right) - \left( a(0) + \frac{b(0)^{3}}{3} \right) = 1 \)
\( \implies a + \frac{b}{3} = 1 \)
\( \implies 3a + b = 3 \) ----- (1) Given that the Expected value \( E(X) = \frac{3}{5} \). The formula for \( E(X) \) is \( \int_{-\infty}^{\infty} x f(x)dx \). So, \( \int_{0}^{1} x(a+bx^{2})dx = \frac{3}{5} \)
\( \implies \int_{0}^{1} (ax+bx^{3})dx = \frac{3}{5} \)
\( \implies \left[ a\frac{x^{2}}{2} + b\frac{x^{4}}{4} \right]_{0}^{1} = \frac{3}{5} \)
\( \implies \left( a\frac{(1)^{2}}{2} + b\frac{(1)^{4}}{4} \right) - \left( a\frac{(0)^{2}}{2} + b\frac{(0)^{4}}{4} \right) = \frac{3}{5} \)
\( \implies \frac{a}{2} + \frac{b}{4} = \frac{3}{5} \) To clear the denominators, multiply by 20 (LCM of 2, 4, 5): \( 10a + 5b = 12 \) ----- (2) Now we have a system of two linear equations: (1) \( 3a + b = 3 \) (2) \( 10a + 5b = 12 \) From equation (1), express \( b \) in terms of \( a \): \( b = 3 - 3a \) Substitute this into equation (2): \( 10a + 5(3 - 3a) = 12 \) \( 10a + 15 - 15a = 12 \) \( -5a = 12 - 15 \) \( -5a = -3 \) \( a = \frac{3}{5} \) Now substitute the value of \( a \) back into \( b = 3 - 3a \): \( b = 3 - 3\left(\frac{3}{5}\right) \) \( b = 3 - \frac{9}{5} \) \( b = \frac{15-9}{5} \) \( b = \frac{6}{5} \) So, the constants are \( a = \frac{3}{5} \) and \( b = \frac{6}{5} \). (ii) To find the Variance, \( Var(X) = E(X^{2}) - [E(X)]^{2} \). First, calculate \( E(X^{2}) = \int_{-\infty}^{\infty} x^{2}f(x)dx \). \( E(X^{2}) = \int_{0}^{1} x^{2}(a+bx^{2})dx \) Substitute \( a = \frac{3}{5} \) and \( b = \frac{6}{5} \): \( E(X^{2}) = \int_{0}^{1} x^{2}\left(\frac{3}{5} + \frac{6}{5}x^{2}\right)dx \) \( E(X^{2}) = \int_{0}^{1} \left(\frac{3}{5}x^{2} + \frac{6}{5}x^{4}\right)dx \)
\( \implies E(X^{2}) = \left[ \frac{3}{5}\frac{x^{3}}{3} + \frac{6}{5}\frac{x^{5}}{5} \right]_{0}^{1} \)
\( \implies E(X^{2}) = \left[ \frac{x^{3}}{5} + \frac{6x^{5}}{25} \right]_{0}^{1} \)
\( \implies E(X^{2}) = \left( \frac{1^{3}}{5} + \frac{6(1)^{5}}{25} \right) - \left( \frac{0^{3}}{5} + \frac{6(0)^{5}}{25} \right) \)
\( \implies E(X^{2}) = \frac{1}{5} + \frac{6}{25} \) To add these, find a common denominator, which is 25: \( E(X^{2}) = \frac{1 \times 5}{5 \times 5} + \frac{6}{25} = \frac{5}{25} + \frac{6}{25} = \frac{11}{25} \) Now calculate \( Var(X) \): \( Var(X) = E(X^{2}) - [E(X)]^{2} \) We found \( E(X^{2}) = \frac{11}{25} \) and we were given \( E(X) = \frac{3}{5} \). \( Var(X) = \frac{11}{25} - \left( \frac{3}{5} \right)^{2} \) \( Var(X) = \frac{11}{25} - \frac{9}{25} \) \( Var(X) = \frac{11-9}{25} \) \( Var(X) = \frac{2}{25} \)
In simple words: For (i), we use two main rules of probability: the total probability over all possible outcomes must add up to 1, and the given expected value. These two rules give us two equations that we solve together to find the unknown values 'a' and 'b'. For (ii), we first calculate the expected value of X squared, then use that along with the given expected value of X to find the variance, which measures how spread out the data is.

🎯 Exam Tip: When solving for multiple unknown constants in a probability density function, you'll often need to set up a system of equations using the total probability rule and the given expected value.

 

Question 6. Prove that if E(X) = 0, then V(X) = E(X²).
Answer:The formula for Variance \( V(X) \) is defined as: \( V(X) = E(X^{2}) - [E(X)]^{2} \) We are given that \( E(X) = 0 \). Substitute this value into the variance formula: \( V(X) = E(X^{2}) - [0]^{2} \) \( V(X) = E(X^{2}) - 0 \) \( V(X) = E(X^{2}) \) Thus, if the expected value of a random variable X is 0, its variance is simply the expected value of X squared. This proves the statement.
In simple words: Variance tells us how spread out the numbers are. The formula for variance involves the average of X squared and the average of X. If the average of X is zero, the formula simplifies, showing that the variance is just the average of X squared. This is a special case that makes calculations easier.

🎯 Exam Tip: Remember the basic formula for variance. If \( E(X) = 0 \), it's a common simplification you can use directly.

 

Question 7. The expected value of a game that works as follows: I flip a coin and, if tails pay you Rs 2; if heads pay you Rs 1. In either case I also pay you Rs 50.
Answer:Let X be the random variable denoting the amount paid for a game of flipping a coin. The possible outcomes for X are: If tails (T), you get Rs 2 + Rs 50 = Rs 52. If heads (H), you get Rs 1 + Rs 50 = Rs 51. The probabilities are: \( P(X=52) = \frac{1}{2} \) (for getting a tail) \( P(X=51) = \frac{1}{2} \) (for getting a head) We can summarize this in a table:

The Expected value \( E(X) \) is calculated as \( \sum_{x} x P(x) \). \( E(X) = (52)\left(\frac{1}{2}\right) + (51)\left(\frac{1}{2}\right) \) \( E(X) = 26 + 25.5 \) \( E(X) = 51.5 \) So, the expected value of the game is Rs 51.5.
In simple words: The expected value is like the average amount you would expect to win if you played the game many times. We calculate it by taking each possible winning amount, multiplying it by its probability, and then adding those results together. Since you get Rs 50 regardless, and either Rs 1 or Rs 2 from the coin flip, the average outcome is Rs 51.5.

🎯 Exam Tip: When calculating expected value for a game, make sure to consider all possible outcomes and their respective probabilities, including any fixed payments.

 

Question 8. Prove that, (i) V(aX) = a²V(X), and (ii) V(X + b) = V(X).
Answer:(i) To prove \( V(aX) = a^{2}V(X) \): We start with the left-hand side (LHS) using the definition of variance: \( LHS = V(aX) \) \( = E((aX)^{2}) - [E(aX)]^{2} \) \( = E(a^{2}X^{2}) - [aE(X)]^{2} \) Since constants can be pulled out of the expectation operator: \( = a^{2}E(X^{2}) - a^{2}[E(X)]^{2} \) Now, factor out \( a^{2} \): \( = a^{2} [E(X^{2}) - [E(X)]^{2}] \) The term inside the square brackets is the definition of \( V(X) \): \( = a^{2}V(X) \) This is the right-hand side (RHS). Therefore, \( V(aX) = a^{2}V(X) \) is proved.
(ii) To prove \( V(X + b) = V(X) \): We start with the left-hand side (LHS) using the definition of variance: \( LHS = V(X + b) \) \( = E((X + b)^{2}) - [E(X + b)]^{2} \) Expand \( (X+b)^{2} \) and \( E(X+b) \): \( E(X+b) = E(X) + E(b) = E(X) + b \) \( E((X+b)^{2}) = E(X^{2} + 2bX + b^{2}) = E(X^{2}) + E(2bX) + E(b^{2}) = E(X^{2}) + 2bE(X) + b^{2} \) Substitute these expanded forms back into the variance equation: \( = (E(X^{2}) + 2bE(X) + b^{2}) - (E(X) + b)^{2} \) \( = E(X^{2}) + 2bE(X) + b^{2} - ([E(X)]^{2} + 2bE(X) + b^{2}) \) Now, remove the parentheses and observe cancellations: \( = E(X^{2}) + 2bE(X) + b^{2} - [E(X)]^{2} - 2bE(X) - b^{2} \) The terms \( 2bE(X) \) and \( b^{2} \) cancel out. \( = E(X^{2}) - [E(X)]^{2} \) This is the definition of \( V(X) \). This is the right-hand side (RHS). Therefore, \( V(X + b) = V(X) \) is proved.
In simple words: For (i), multiplying a random variable by a constant 'a' makes the variance 'a squared' times bigger, because variance measures spread, and squaring 'a' reflects how the spread changes. For (ii), adding a constant 'b' to a random variable does not change its variance, because adding 'b' just shifts all the values, but their spread or distance from each other remains the same.

🎯 Exam Tip: Remember these two key properties of variance: constants multiplying a variable get squared, and constants added to a variable have no effect on variance. These help simplify many problems.

 

Question 9. Consider a random variable X with p.d.f.\[ f(x)= \begin{cases} 3x^{2}, & \text{if } 0 < x < 1 \\ 0, & \text{otherwise} \end{cases} \]Find the expected life of this piece of equipment.
Answer:The expected life of the equipment is the expected value of the random variable X, which is \( E(X) \). For a continuous random variable with probability density function \( f(x) \), \( E(X) = \int_{-\infty}^{\infty} x f(x)dx \). Given \( f(x) = 3x^{2} \) for \( 0 < x < 1 \), we integrate from 0 to 1: \( E(X) = \int_{0}^{1} x (3x^{2})dx \) \( E(X) = \int_{0}^{1} 3x^{3}dx \) Now, integrate the expression: \( E(X) = \left[ 3 \frac{x^{4}}{4} \right]_{0}^{1} \)
\( \implies E(X) = \frac{3}{4} [x^{4}]_{0}^{1} \)
\( \implies E(X) = \frac{3}{4} (1^{4} - 0^{4}) \)
\( \implies E(X) = \frac{3}{4} (1 - 0) \)
\( \implies E(X) = \frac{3}{4} \) So, the expected life of the equipment is \( \frac{3}{4} \) (in thousands of hours, if that's the unit for X). We can also compute \( E(X^2) \) and \( Var(X) \) as a bonus, since this is a typical follow-up. \( E(X^{2}) = \int_{0}^{1} x^{2} (3x^{2})dx \) \( E(X^{2}) = \int_{0}^{1} 3x^{4}dx \) \( E(X^{2}) = \left[ 3 \frac{x^{5}}{5} \right]_{0}^{1} \) \( E(X^{2}) = \frac{3}{5} [x^{5}]_{0}^{1} \) \( E(X^{2}) = \frac{3}{5} (1^{5} - 0^{5}) \) \( E(X^{2}) = \frac{3}{5} \) Now for \( Var(X) \): \( Var(X) = E(X^{2}) - [E(X)]^{2} \) \( Var(X) = \frac{3}{5} - \left( \frac{3}{4} \right)^{2} \) \( Var(X) = \frac{3}{5} - \frac{9}{16} \) To subtract these, find a common denominator, which is 80: \( Var(X) = \frac{3 \times 16}{5 \times 16} - \frac{9 \times 5}{16 \times 5} \) \( Var(X) = \frac{48}{80} - \frac{45}{80} \) \( Var(X) = \frac{3}{80} \) If we need to find \( V(3X-2) \): \( V(aX+b) = a^2 V(X) \) \( V(3X-2) = (3)^{2} Var(X) \) \( V(3X-2) = 9 \left( \frac{3}{80} \right) \) \( V(3X-2) = \frac{27}{80} \)
In simple words: The "expected life" of the equipment means the average time it is expected to work. We find this by using a mathematical tool called integration. We take the formula for the probability of the equipment working for a certain time, multiply it by the time itself, and then sum this up over all possible times. This gives us the average lifespan.

🎯 Exam Tip: For continuous random variables, the expected value is found by integrating \( x \cdot f(x) \) over the entire range of the variable. Do not forget to include the 'x' in the integrand.

 

Question 10. The time to failure in thousands of hours of an important piece of electronic equipment, a manufactured DVD player has the density function\[ f(x)= \begin{cases} 2e^{-2x}, & x>0 \\ 0, & \text{otherwise} \end{cases} \]Find the expected life of this piece of equipment.
Answer:The expected life of the equipment is the expected value of the random variable X, which is \( E(X) \). For a continuous random variable with probability density function \( f(x) \), \( E(X) = \int_{-\infty}^{\infty} x f(x)dx \). Given \( f(x) = 2e^{-2x} \) for \( x > 0 \), we integrate from 0 to infinity: \( E(X) = \int_{0}^{\infty} x (2e^{-2x})dx \) \( E(X) = 2 \int_{0}^{\infty} xe^{-2x}dx \) This integral is a form of the Gamma function, specifically related to the definition of \( \int_{0}^{\infty} x^{n}e^{-ax}dx = \frac{n!}{a^{n+1}} \). Here, \( n = 1 \) and \( a = 2 \). So, \( \int_{0}^{\infty} xe^{-2x}dx = \frac{1!}{2^{1+1}} = \frac{1}{2^{2}} = \frac{1}{4} \) Now, substitute this back into the expression for \( E(X) \): \( E(X) = 2 \left( \frac{1}{4} \right) \) \( E(X) = \frac{1}{2} \) So, the expected life of the equipment is \( \frac{1}{2} \) thousand hours, which means 500 hours. This exponential distribution is commonly used for equipment lifetimes.
In simple words: The expected life is the average time we expect the DVD player to last. We find this by using a special mathematical process called integration on its density function. The density function tells us the probability of failure at different times. By calculating this integral, we get a single number that represents its average lifespan.

🎯 Exam Tip: Recognize common integral forms like \( \int_{0}^{\infty} x^{n}e^{-ax}dx = \frac{n!}{a^{n+1}} \) (Gamma function related) to quickly solve expected value problems for exponential distributions.