Samacheer Kalvi Class 12 Maths Solutions Chapter 11 Probability Distributions Exercise 11.3

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 11 Probability Distributions Ex 11.3

Chapter 1

Question 1. The probability density function of X is given by
\[ f(x) = \begin{cases} ke^{-x/3} & \text{for } x > 0 \\ 0 & \text{for } x \le 0 \end{cases} \]
Find the value of k.
Answer: Since \( f(x) \) is a probability density function, the total probability over all possible values must be equal to 1. This means the integral of \( f(x) \) from negative infinity to positive infinity is 1. We only need to integrate from 0 to infinity because for \( x \le 0 \), the function \( f(x) = 0 \). Setting up the integral and solving for k, we find that the value of k is 1/3, which is a common value in many exponential distributions.
\[ \int_{-\infty}^{\infty} f(x)dx = 1 \]
\[ \int_{-\infty}^{0} f(x)dx + \int_{0}^{\infty} f(x)dx = 1 \]
\[ 0 + \int_{0}^{\infty} k e^{-x/3}dx = 1 \]
\[ k \left[ \frac{e^{-x/3}}{-1/3} \right]_{0}^{\infty} = 1 \]
\[ k [-3e^{-x/3}]_{0}^{\infty} = 1 \]
\[ k [(-3e^{-\infty}) - (-3e^{0})] = 1 \]
\[ k [0 - (-3)] = 1 \]
\[ 3k = 1 \]
\[ k = \frac{1}{3} \]
In simple words: A probability function must add up to 1 when you look at all possible outcomes. We calculate the unknown value 'k' by doing this sum (called integration) from 0 to infinity, which gives us k equals 1/3.

🎯 Exam Tip: Remember that for any probability density function, the integral over its entire domain must always equal 1. This is a fundamental property used to find unknown constants.

Question 2. The probability density function of X is given by
\[ f(x)= \begin{cases} x & \text{ for } 0 \le x < 1 \\ 2-x & \text{ for } 1 \le x < 2 \\ 0 & \text{ otherwise } \end{cases} \]
Find
(i) P(0.2 ≤ X < 0.6)
(ii) P(1.2 ≤ X < 1.8)
(iii) P(0.5 ≤ X < 1.5)
Answer: To find the probability over a certain range for a given probability density function, we need to integrate the function within that specific range. We do this for each sub-part, breaking down the integral if the range spans across different definitions of \( f(x) \).
(i) \( P(0.2 \le X < 0.6) = \int_{0.2}^{0.6} f(x)dx \)
Since \( 0.2 \le X < 0.6 \) falls within the range \( 0 \le x < 1 \), where \( f(x) = x \), we integrate \( x \):
\( \int_{0.2}^{0.6} x dx = \left[ \frac{x^2}{2} \right]_{0.2}^{0.6} \)
\( = \frac{(0.6)^2}{2} - \frac{(0.2)^2}{2} \)
\( = \frac{0.36}{2} - \frac{0.04}{2} \)
\( = 0.18 - 0.02 = 0.16 \)

(ii) \( P(1.2 \le X < 1.8) = \int_{1.2}^{1.8} f(x)dx \)
Since \( 1.2 \le X < 1.8 \) falls within the range \( 1 \le x < 2 \), where \( f(x) = 2-x \), we integrate \( 2-x \):
\( \int_{1.2}^{1.8} (2-x) dx = \left[ 2x - \frac{x^2}{2} \right]_{1.2}^{1.8} \)
\( = \left( 2(1.8) - \frac{(1.8)^2}{2} \right) - \left( 2(1.2) - \frac{(1.2)^2}{2} \right) \)
\( = \left( 3.6 - \frac{3.24}{2} \right) - \left( 2.4 - \frac{1.44}{2} \right) \)
\( = (3.6 - 1.62) - (2.4 - 0.72) \)
\( = 1.98 - 1.68 = 0.3 \)

(iii) \( P(0.5 \le X < 1.5) = \int_{0.5}^{1.5} f(x)dx \)
This range covers two different definitions of \( f(x) \): \( 0.5 \le x < 1 \) (where \( f(x)=x \)) and \( 1 \le x < 1.5 \) (where \( f(x)=2-x \)). So we split the integral:
\( P(0.5 \le X < 1.5) = \int_{0.5}^{1} f(x)dx + \int_{1}^{1.5} f(x)dx \)
\( = \int_{0.5}^{1} x dx + \int_{1}^{1.5} (2-x) dx \)
\( = \left[ \frac{x^2}{2} \right]_{0.5}^{1} + \left[ 2x - \frac{x^2}{2} \right]_{1}^{1.5} \)
\( = \left( \frac{1^2}{2} - \frac{(0.5)^2}{2} \right) + \left( \left( 2(1.5) - \frac{(1.5)^2}{2} \right) - \left( 2(1) - \frac{1^2}{2} \right) \right) \)
\( = \left( \frac{1}{2} - \frac{0.25}{2} \right) + \left( \left( 3 - \frac{2.25}{2} \right) - \left( 2 - \frac{1}{2} \right) \right) \)
\( = (0.5 - 0.125) + ((3 - 1.125) - (2 - 0.5)) \)
\( = 0.375 + (1.875 - 1.5) \)
\( = 0.375 + 0.375 = 0.75 \)
In simple words: To find the probability over a range, you calculate the area under the curve of the function for that range. If the function changes its rule within that range, you break the calculation into parts and add them up.

🎯 Exam Tip: When the probability density function is defined in parts, be careful to split your integral at the points where the function definition changes. Always ensure your limits of integration match the function part being used.

Question 3. Suppose the amount of milk sold daily at a milk booth is distributed with a minimum of 200 litres and a maximum of 600 litres with probability density function
\[ f(x) = \begin{cases} k & \text{ for } 200 \le x \le 600 \\ 0 & \text{ otherwise } \end{cases} \]
Find
(i) the value of k
(ii) the distribution function
(iii) the probability that daily sales will fall between 300 litres and 500 litres?
Answer: For a probability density function, the total area under its curve must be equal to 1. We use this property to find the unknown constant 'k'. The distribution function \( F(x) \) is the cumulative probability, found by integrating \( f(x) \) from negative infinity up to \( x \).
(i) To find the value of k:
Since \( f(x) \) is a probability density function, \( \int_{-\infty}^{\infty} f(x)dx = 1 \).
\[ \int_{-\infty}^{200} f(x)dx + \int_{200}^{600} f(x)dx + \int_{600}^{\infty} f(x)dx = 1 \]
\[ 0 + \int_{200}^{600} k dx + 0 = 1 \]
\[ k[x]_{200}^{600} = 1 \]
\[ k(600 - 200) = 1 \]
\[ k(400) = 1 \]
\[ k = \frac{1}{400} \]
(ii) To find the distribution function \( F(x) = \int_{-\infty}^{x} f(u)du \):

Case 1: For \( x < 200 \)
\( F(x) = \int_{-\infty}^{x} f(u)du = 0 \)
(Since \( f(u) = 0 \) for \( u < 200 \))

Case 2: For \( 200 \le x \le 600 \)
\( F(x) = \int_{-\infty}^{x} f(u)du = \int_{-\infty}^{200} f(u)du + \int_{200}^{x} f(u)du \)
\( = 0 + \int_{200}^{x} k du \)
\( = k[u]_{200}^{x} = k(x - 200) \)
Substitute \( k = \frac{1}{400} \):
\( F(x) = \frac{1}{400}(x - 200) \)
\( F(x) = \frac{x}{400} - \frac{200}{400} = \frac{x}{400} - \frac{1}{2} \)

Case 3: For \( x > 600 \)
\( F(x) = \int_{-\infty}^{x} f(u)du = \int_{-\infty}^{200} f(u)du + \int_{200}^{600} f(u)du + \int_{600}^{x} f(u)du \)
\( = 0 + \int_{200}^{600} k du + 0 \)
\( = k[u]_{200}^{600} = k(600 - 200) \)
\( = \frac{1}{400}(400) = 1 \)

Combining these cases, the distribution function is:
\[ F(x) = \begin{cases} 0 & \text{for } x < 200 \\ \frac{x}{400} - \frac{1}{2} & \text{for } 200 \le x \le 600 \\ 1 & \text{for } x > 600 \end{cases} \]
(iii) To find the probability that daily sales will fall between 300 litres and 500 litres:
\( P(300 < X < 500) = \int_{300}^{500} f(x)dx \)
Since \( 300 \le x \le 500 \) is within the range where \( f(x) = k \), we integrate \( k \):
\( = \int_{300}^{500} k dx = k[x]_{300}^{500} \)
\( = k(500 - 300) = k(200) \)
Substitute \( k = \frac{1}{400} \):
\( = \frac{1}{400} \times 200 = \frac{200}{400} = \frac{1}{2} \)
In simple words: We first found 'k' by making sure the total chance is 1. Then we made a cumulative function \( F(x) \) which shows the chance of getting a value up to 'x'. Finally, we found the chance of sales being between 300 and 500 by integrating \( f(x) \) over that range. This means there is a 50% chance sales will be in that range.

🎯 Exam Tip: When dealing with uniform probability distributions (where \( f(x) \) is a constant k over an interval), remembering the formula \( k = \frac{1}{\text{range}} \) can save time. The distribution function \( F(x) \) is essential for calculating probabilities over any interval using \( P(a < X < b) = F(b) - F(a) \).

Question 4. The probability density function of X is given by
\[ f(x) = \begin{cases} ke^{-x/3} & \text{for } x > 0 \\ 0 & \text{for } x \le 0 \end{cases} \]
(i) the value of k
(ii) the distribution function
(iii) P(X < 3)
(iv) P(5 ≤ X)
(v) P(X ≤ 4)
Answer: We need to determine the value of 'k' that ensures the given function is a valid probability density function, meaning its total integral over all possible values is 1. Then, we find the cumulative distribution function and calculate probabilities for various ranges using integration.
(i) To find the value of k:
Since \( f(x) \) is a probability density function, \( \int_{-\infty}^{\infty} f(x)dx = 1 \).
\[ \int_{-\infty}^{0} f(x)dx + \int_{0}^{\infty} f(x)dx = 1 \]
\[ 0 + \int_{0}^{\infty} k e^{-x/3}dx = 1 \]
\[ k \left[ \frac{e^{-x/3}}{-1/3} \right]_{0}^{\infty} = 1 \]
\[ k [-3e^{-x/3}]_{0}^{\infty} = 1 \]
\[ k [(-3e^{-\infty}) - (-3e^{0})] = 1 \]
\[ k [0 - (-3)] = 1 \]
\[ 3k = 1 \]
\[ k = \frac{1}{3} \]
(ii) To find the distribution function \( F(x) = \int_{-\infty}^{x} f(u)du \):

Case 1: For \( x < 0 \)
\( F(x) = \int_{-\infty}^{x} f(u)du = 0 \)
(Since \( f(u) = 0 \) for \( u < 0 \))

Case 2: For \( x \ge 0 \)
\( F(x) = \int_{-\infty}^{x} f(u)du = \int_{-\infty}^{0} f(u)du + \int_{0}^{x} f(u)du \)
\( = 0 + \int_{0}^{x} k e^{-u/3}du \)
Substitute \( k = \frac{1}{3} \):
\( = \frac{1}{3} \int_{0}^{x} e^{-u/3}du \)
\( = \frac{1}{3} \left[ \frac{e^{-u/3}}{-1/3} \right]_{0}^{x} \)
\( = \frac{1}{3} (-3) [e^{-u/3}]_{0}^{x} \)
\( = -1 [e^{-x/3} - e^{0}] \)
\( = -(e^{-x/3} - 1) = 1 - e^{-x/3} \)

Combining these cases, the distribution function is:
\[ F(x) = \begin{cases} 0 & \text{for } x < 0 \\ 1 - e^{-x/3} & \text{for } x \ge 0 \end{cases} \]
(iii) To find \( P(X < 3) \):
\( P(X < 3) = F(3) = 1 - e^{-3/3} = 1 - e^{-1} \)
Alternatively, using integration:
\( P(X < 3) = \int_{0}^{3} f(x)dx = \int_{0}^{3} k e^{-x/3}dx \)
\( = \frac{1}{3} \int_{0}^{3} e^{-x/3}dx \)
\( = \frac{1}{3} \left[ \frac{e^{-x/3}}{-1/3} \right]_{0}^{3} \)
\( = -1 [e^{-3/3} - e^{0}] \)
\( = -(e^{-1} - 1) = 1 - e^{-1} \)
(iv) To find \( P(5 \le X) \):
\( P(5 \le X) = 1 - P(X < 5) = 1 - F(5) \)
\( = 1 - (1 - e^{-5/3}) = e^{-5/3} \)
Alternatively, using integration:
\( P(5 \le X) = \int_{5}^{\infty} f(x)dx = \int_{5}^{\infty} k e^{-x/3}dx \)
\( = \frac{1}{3} \int_{5}^{\infty} e^{-x/3}dx \)
\( = \frac{1}{3} \left[ \frac{e^{-x/3}}{-1/3} \right]_{5}^{\infty} \)
\( = -1 [e^{-\infty} - e^{-5/3}] \)
\( = -(0 - e^{-5/3}) = e^{-5/3} \)
(v) To find \( P(X \le 4) \):
\( P(X \le 4) = F(4) = 1 - e^{-4/3} \)
Alternatively, using integration:
\( P(X \le 4) = \int_{0}^{4} f(x)dx = \int_{0}^{4} k e^{-x/3}dx \)
\( = \frac{1}{3} \int_{0}^{4} e^{-x/3}dx \)
\( = \frac{1}{3} \left[ \frac{e^{-x/3}}{-1/3} \right]_{0}^{4} \)
\( = -1 [e^{-4/3} - e^{0}] \)
\( = -(e^{-4/3} - 1) = 1 - e^{-4/3} \)
In simple words: First, we calculated the constant 'k' so that the total probability is 1. Then we built the cumulative distribution function, which tells us the chance of a value being less than or equal to a certain point. Finally, we used this cumulative function to quickly find the probabilities for different ranges.

🎯 Exam Tip: For exponential distributions, remembering the cumulative distribution function \( F(x) = 1 - e^{-\lambda x} \) for \( x \ge 0 \) (where \( \lambda = 1/3 \) in this case) can significantly speed up probability calculations like \( P(X < a) = F(a) \) and \( P(X \ge a) = 1 - F(a) \).

Question 5. If X is the random variable with probability density function f(x) given by,
\[ f(x)= \begin{cases} x+1 & \text{ for } -1 \le x < 0 \\ -x+1 & \text{ for } 0 \le x < 1 \\ 0 & \text{ otherwise } \end{cases} \]
(i) the distribution function F(x)
(ii) P(-0.5 ≤ x ≤ 0.5)
Answer: We need to find the cumulative distribution function \( F(x) \) by integrating the probability density function \( f(x) \) from negative infinity up to \( x \), considering its different definitions in parts. Then, we use \( F(x) \) to calculate the probability over a specific interval.
(i) To find the distribution function \( F(x) = \int_{-\infty}^{x} f(u)du \):

Case 1: For \( x < -1 \)
\( F(x) = \int_{-\infty}^{x} f(u)du = 0 \)
(Since \( f(u) = 0 \) for \( u < -1 \))

Case 2: For \( -1 \le x < 0 \)
\( F(x) = \int_{-\infty}^{x} f(u)du = \int_{-\infty}^{-1} f(u)du + \int_{-1}^{x} f(u)du \)
\( = 0 + \int_{-1}^{x} (u+1)du \)
\( = \left[ \frac{u^2}{2} + u \right]_{-1}^{x} \)
\( = \left( \frac{x^2}{2} + x \right) - \left( \frac{(-1)^2}{2} + (-1) \right) \)
\( = \frac{x^2}{2} + x - \left( \frac{1}{2} - 1 \right) \)
\( = \frac{x^2}{2} + x - \left( -\frac{1}{2} \right) \)
\( = \frac{x^2}{2} + x + \frac{1}{2} \)

Case 3: For \( 0 \le x < 1 \)
\( F(x) = \int_{-\infty}^{x} f(u)du = \int_{-\infty}^{-1} f(u)du + \int_{-1}^{0} f(u)du + \int_{0}^{x} f(u)du \)
\( = 0 + \left[ \frac{u^2}{2} + u \right]_{-1}^{0} + \int_{0}^{x} (-u+1)du \)
\( = \left( \frac{0^2}{2} + 0 \right) - \left( \frac{(-1)^2}{2} + (-1) \right) + \left[ -\frac{u^2}{2} + u \right]_{0}^{x} \)
\( = (0 - (\frac{1}{2} - 1)) + \left( \left( -\frac{x^2}{2} + x \right) - \left( -\frac{0^2}{2} + 0 \right) \right) \)
\( = -\left( -\frac{1}{2} \right) + \left( -\frac{x^2}{2} + x \right) \)
\( = \frac{1}{2} - \frac{x^2}{2} + x \)

Case 4: For \( x \ge 1 \)
\( F(x) = \int_{-\infty}^{x} f(u)du = \int_{-\infty}^{-1} f(u)du + \int_{-1}^{0} f(u)du + \int_{0}^{1} f(u)du + \int_{1}^{x} f(u)du \)
\( = 0 + \left[ \frac{u^2}{2} + u \right]_{-1}^{0} + \left[ -\frac{u^2}{2} + u \right]_{0}^{1} + 0 \)
\( = \left( 0 - \left( \frac{1}{2} - 1 \right) \right) + \left( \left( -\frac{1}{2} + 1 \right) - 0 \right) \)
\( = -\left( -\frac{1}{2} \right) + \frac{1}{2} \)
\( = \frac{1}{2} + \frac{1}{2} = 1 \)

Combining these cases, the distribution function is:
\[ F(x) = \begin{cases} 0 & \text{for } x < -1 \\ \frac{x^2}{2} + x + \frac{1}{2} & \text{for } -1 \le x < 0 \\ -\frac{x^2}{2} + x + \frac{1}{2} & \text{for } 0 \le x < 1 \\ 1 & \text{for } x \ge 1 \end{cases} \]
(ii) To find \( P(-0.5 \le X \le 0.5) \):
\( P(-0.5 \le X \le 0.5) = F(0.5) - F(-0.5) \)
Using the definition of \( F(x) \):
For \( F(0.5) \), since \( 0 \le 0.5 < 1 \), we use \( F(x) = -\frac{x^2}{2} + x + \frac{1}{2} \):
\( F(0.5) = -\frac{(0.5)^2}{2} + (0.5) + \frac{1}{2} = -\frac{0.25}{2} + 0.5 + 0.5 = -0.125 + 1 = 0.875 \)
For \( F(-0.5) \), since \( -1 \le -0.5 < 0 \), we use \( F(x) = \frac{x^2}{2} + x + \frac{1}{2} \):
\( F(-0.5) = \frac{(-0.5)^2}{2} + (-0.5) + \frac{1}{2} = \frac{0.25}{2} - 0.5 + 0.5 = 0.125 \)
Therefore,
\( P(-0.5 \le X \le 0.5) = F(0.5) - F(-0.5) = 0.875 - 0.125 = 0.75 \)
Alternatively, using integration:
\( P(-0.5 \le X \le 0.5) = \int_{-0.5}^{0.5} f(x)dx \)
This integral needs to be split at \( x=0 \) because the function definition changes:
\( P(-0.5 \le X \le 0.5) = \int_{-0.5}^{0} (x+1)dx + \int_{0}^{0.5} (-x+1)dx \)
\( = \left[ \frac{x^2}{2} + x \right]_{-0.5}^{0} + \left[ -\frac{x^2}{2} + x \right]_{0}^{0.5} \)
\( = \left( \frac{0^2}{2} + 0 \right) - \left( \frac{(-0.5)^2}{2} + (-0.5) \right) + \left( -\frac{(0.5)^2}{2} + 0.5 \right) - \left( -\frac{0^2}{2} + 0 \right) \)
\( = (0 - (\frac{0.25}{2} - 0.5)) + (-\frac{0.25}{2} + 0.5 - 0) \)
\( = -(0.125 - 0.5) + (-0.125 + 0.5) \)
\( = -(-0.375) + 0.375 \)
\( = 0.375 + 0.375 = 0.75 \)
In simple words: We built the cumulative distribution function \( F(x) \) by adding up the probabilities from the left, step by step, for each part of the function. Then, to find the probability within a range, we simply subtracted the \( F(x) \) value at the lower limit from the \( F(x) \) value at the upper limit. This gives us the total chance of the variable falling within that specific range.

🎯 Exam Tip: When finding \( F(x) \) for piecewise functions, remember to evaluate the previous integral at its upper limit before starting the next part, ensuring continuity. Also, check that \( F(x) \) is always non-decreasing and starts at 0 and ends at 1.

Question 6. If X is the random variable with distribution function F(x) given by,
\[ F(x) = \begin{cases} 0 & \text{ for } x < 0 \\ \frac{1}{2}(x^2+x) & \text{ for } 0 \le x < 1 \\ 1 & \text{ for } x \ge 1 \end{cases} \]
Find
(i) the probability density function f(x)
(ii) P(0.3 ≤ X ≤ 0.6)
Answer: We are given the cumulative distribution function \( F(x) \) and need to find the probability density function \( f(x) \) by differentiating \( F(x) \). Then, we use the properties of \( F(x) \) to calculate the probability over a specified interval.
(i) To find the probability density function \( f(x) \):
The probability density function \( f(x) \) is the derivative of the distribution function \( F(x) \), i.e., \( f(x) = F'(x) \).
For \( x < 0 \): \( F(x) = 0 \implies f(x) = 0 \)
For \( 0 \le x < 1 \): \( F(x) = \frac{1}{2}(x^2+x) \)
\( f(x) = \frac{d}{dx} \left( \frac{1}{2}(x^2+x) \right) = \frac{1}{2}(2x+1) = x + \frac{1}{2} \)
For \( x \ge 1 \): \( F(x) = 1 \implies f(x) = 0 \)
So, the probability density function is:
\[ f(x) = \begin{cases} 0 & \text{ for } x < 0 \\ x + \frac{1}{2} & \text{ for } 0 \le x < 1 \\ 0 & \text{ otherwise } \end{cases} \]
(ii) To find \( P(0.3 \le X \le 0.6) \):
\( P(0.3 \le X \le 0.6) = F(0.6) - F(0.3) \)
Using the definition of \( F(x) \) for \( 0 \le x < 1 \), which is \( F(x) = \frac{1}{2}(x^2+x) \):
\( F(0.6) = \frac{1}{2}((0.6)^2 + 0.6) = \frac{1}{2}(0.36 + 0.6) = \frac{1}{2}(0.96) = 0.48 \)
\( F(0.3) = \frac{1}{2}((0.3)^2 + 0.3) = \frac{1}{2}(0.09 + 0.3) = \frac{1}{2}(0.39) = 0.195 \)
Therefore,
\( P(0.3 \le X \le 0.6) = F(0.6) - F(0.3) = 0.48 - 0.195 = 0.285 \)
In simple words: The probability density function \( f(x) \) is found by taking the derivative of the cumulative distribution function \( F(x) \). Once we have \( F(x) \), finding the probability for a range is as simple as subtracting the \( F \) value at the start of the range from the \( F \) value at the end.

🎯 Exam Tip: Remember the inverse relationship: differentiating \( F(x) \) gives \( f(x) \), and integrating \( f(x) \) gives \( F(x) \). When calculating probabilities for a range using \( F(x) \), always use \( P(a \le X \le b) = F(b) - F(a) \).

Chapter 1

Question 1. The probability density function of X is given by
\[ f(x) = \begin{cases} ke^{-x/3} & \text{for } x > 0 \\ 0 & \text{for } x \le 0 \end{cases} \]
Find the value of k.
Answer: Since \( f(x) \) is a probability density function, the total probability over all possible values must be 1. This means the integral of \( f(x) \) from \( -\infty \) to \( \infty \) should equal 1. We split the integral into two parts based on the definition of \( f(x) \).
\[ \int_{-\infty}^{\infty} f(x)dx = 1 \]
\[ \int_{-\infty}^{0} f(x)dx + \int_{0}^{\infty} f(x)dx = 1 \]
From the definition, \( f(x) = 0 \) for \( x \le 0 \), so the first integral is 0. We only need to integrate \( ke^{-x/3} \) from 0 to \( \infty \).
\[ 0 + \int_{0}^{\infty} ke^{-x/3}dx = 1 \]
To integrate \( e^{-x/3} \), we can use a substitution or direct integration. The integral of \( e^{ax} \) is \( \frac{1}{a}e^{ax} \). Here \( a = -\frac{1}{3} \).
\[ k \left[ \frac{e^{-x/3}}{-1/3} \right]_{0}^{\infty} = 1 \]
\[ k \left[ -3e^{-x/3} \right]_{0}^{\infty} = 1 \]
Now we evaluate the limits. As \( x \to \infty \), \( e^{-x/3} \to 0 \). When \( x = 0 \), \( e^{0} = 1 \).
\[ k (-3e^{-\infty} - (-3e^{0})) = 1 \]
\[ k (0 - (-3 \times 1)) = 1 \]
\[ k (3) = 1 \]
\[ k = \frac{1}{3} \]
Thus, the value of \( k \) is \( \frac{1}{3} \). This constant ensures that the total area under the probability density curve is 1, a fundamental property of such functions.
In simple words: A probability density function must have an area of 1 when you integrate it over all possible values. We split the integral into two parts, but one part is zero because the function is defined as zero for negative x. We then solve the integral for the positive x range to find the value of 'k' that makes the total area exactly 1.

🎯 Exam Tip: Remember that for any probability density function, the total area under its curve over its entire domain must always be equal to 1. This is a key property used to find unknown constants like 'k'.

Question 2. The probability density function of X is
\[ f(x) = \begin{cases} x & \text{for } 0 \le x < 1 \\ 2-x & \text{for } 1 \le x < 2 \\ 0 & \text{otherwise} \end{cases} \]
Find
(i) \( P(0.2 \le X < 0.6) \)
(ii) \( P(1.2 \le X < 1.8) \)
(iii) \( P(0.5 \le X < 1.5) \)
Answer: To find the probability over a certain range, we need to integrate the probability density function \( f(x) \) over that range. The function changes its definition at \( x=1 \), so we must pay attention to this when integrating.
(i) For \( P(0.2 \le X < 0.6) \), the range \( [0.2, 0.6) \) falls entirely within \( 0 \le x < 1 \), where \( f(x) = x \).
\[ P(0.2 \le X < 0.6) = \int_{0.2}^{0.6} f(x)dx \]
\[ = \int_{0.2}^{0.6} x dx \]
\[ = \left[ \frac{x^2}{2} \right]_{0.2}^{0.6} \]
\[ = \frac{(0.6)^2}{2} - \frac{(0.2)^2}{2} \]
\[ = \frac{0.36}{2} - \frac{0.04}{2} \]
\[ = 0.18 - 0.02 = 0.16 \]
(ii) For \( P(1.2 \le X < 1.8) \), the range \( [1.2, 1.8) \) falls entirely within \( 1 \le x < 2 \), where \( f(x) = 2-x \).
\[ P(1.2 \le X < 1.8) = \int_{1.2}^{1.8} f(x)dx \]
\[ = \int_{1.2}^{1.8} (2-x) dx \]
\[ = \left[ 2x - \frac{x^2}{2} \right]_{1.2}^{1.8} \]
\[ = \left( 2(1.8) - \frac{(1.8)^2}{2} \right) - \left( 2(1.2) - \frac{(1.2)^2}{2} \right) \]
\[ = \left( 3.6 - \frac{3.24}{2} \right) - \left( 2.4 - \frac{1.44}{2} \right) \]
\[ = (3.6 - 1.62) - (2.4 - 0.72) \]
\[ = 1.98 - 1.68 = 0.3 \]
(iii) For \( P(0.5 \le X < 1.5) \), the range \( [0.5, 1.5) \) crosses the point where the function definition changes (at \( x=1 \)). We must split the integral into two parts: from 0.5 to 1, and from 1 to 1.5.
\[ P(0.5 \le X < 1.5) = \int_{0.5}^{1.5} f(x)dx \]
\[ = \int_{0.5}^{1} f(x)dx + \int_{1}^{1.5} f(x)dx \]
For \( 0.5 \le x < 1 \), \( f(x) = x \). For \( 1 \le x < 1.5 \), \( f(x) = 2-x \).
\[ = \int_{0.5}^{1} x dx + \int_{1}^{1.5} (2-x) dx \]
\[ = \left[ \frac{x^2}{2} \right]_{0.5}^{1} + \left[ 2x - \frac{x^2}{2} \right]_{1}^{1.5} \]
\[ = \left( \frac{1^2}{2} - \frac{(0.5)^2}{2} \right) + \left( \left( 2(1.5) - \frac{(1.5)^2}{2} \right) - \left( 2(1) - \frac{1^2}{2} \right) \right) \]
\[ = \left( \frac{1}{2} - \frac{0.25}{2} \right) + \left( \left( 3 - \frac{2.25}{2} \right) - \left( 2 - \frac{1}{2} \right) \right) \]
\[ = (0.5 - 0.125) + ( (3 - 1.125) - (2 - 0.5) ) \]
\[ = 0.375 + (1.875 - 1.5) \]
\[ = 0.375 + 0.375 = 0.75 \]
The total probability is 0.75. Understanding how to split the integral at points where the function definition changes is crucial for these types of problems.
In simple words: To find the chance of something happening within a range, we add up all the little probabilities over that range using integration. If the rule for the probability function changes at some point, we need to break our calculation into parts at that point and add them up.

🎯 Exam Tip: Always check the definition of \( f(x) \) and split the integral whenever the given interval crosses a point where the function's rule changes. This ensures you use the correct expression for \( f(x) \) in each segment.

Question 3. Suppose the amount of milk sold daily at a milk booth is distributed with a minimum of 200 litres and a maximum of 600 litres with probability density function
\[ f(x) = \begin{cases} k & \text{for } 200 \le x \le 600 \\ 0 & \text{otherwise} \end{cases} \]
Find
(i) the value of k
(ii) the distribution function
(iii) the probability that daily sales will fall between 300 litres and 500 litres?
Answer: This problem involves a continuous uniform distribution, where the probability density is constant over a given range and zero elsewhere.
(i) To find the value of \( k \), we use the property that the total probability must be 1. We integrate \( f(x) \) over its entire domain.
\[ \int_{-\infty}^{\infty} f(x)dx = 1 \]
\[ \int_{-\infty}^{200} f(x)dx + \int_{200}^{600} f(x)dx + \int_{600}^{\infty} f(x)dx = 1 \]
Since \( f(x) = 0 \) outside the range \( [200, 600] \), the first and third integrals are zero.
\[ 0 + \int_{200}^{600} k dx + 0 = 1 \]
\[ k [x]_{200}^{600} = 1 \]
\[ k (600 - 200) = 1 \]
\[ k (400) = 1 \]
\[ k = \frac{1}{400} \]
(ii) The distribution function, \( F(x) \), is the cumulative probability up to a certain point \( x \). It is defined as \( F(x) = \int_{-\infty}^{x} f(u)du \). We need to consider three cases:
**Case 1: \( x < 200 \)**
Since \( f(u) = 0 \) for \( u < 200 \), the integral is 0.
\[ F(x) = \int_{-\infty}^{x} f(u)du = 0 \]
**Case 2: \( 200 \le x \le 600 \)**
Here, we integrate from 200 up to \( x \), using \( f(u) = k = \frac{1}{400} \).
\[ F(x) = \int_{-\infty}^{x} f(u)du = \int_{-\infty}^{200} f(u)du + \int_{200}^{x} f(u)du \]
\[ = 0 + \int_{200}^{x} \frac{1}{400} du \]
\[ = \frac{1}{400} [u]_{200}^{x} \]
\[ = \frac{1}{400} (x - 200) \]
**Case 3: \( x > 600 \)**
At this point, all the probability has accumulated, so \( F(x) \) must be 1.
\[ F(x) = \int_{-\infty}^{x} f(u)du = \int_{-\infty}^{200} f(u)du + \int_{200}^{600} f(u)du + \int_{600}^{x} f(u)du \]
\[ = 0 + \int_{200}^{600} \frac{1}{400} du + 0 \]
\[ = \frac{1}{400} [u]_{200}^{600} \]
\[ = \frac{1}{400} (600 - 200) \]
\[ = \frac{1}{400} (400) = 1 \]
Combining these cases, the distribution function \( F(x) \) is:
\[ F(x) = \begin{cases} 0 & \text{for } x < 200 \\ \frac{x-200}{400} & \text{for } 200 \le x \le 600 \\ 1 & \text{for } x > 600 \end{cases} \]
(iii) To find the probability that daily sales will fall between 300 litres and 500 litres, \( P(300 < X < 500) \), we integrate \( f(x) \) from 300 to 500. Both 300 and 500 are within the range \( [200, 600] \), so \( f(x) = k = \frac{1}{400} \).
\[ P(300 < X < 500) = \int_{300}^{500} f(x)dx \]
\[ = \int_{300}^{500} \frac{1}{400} dx \]
\[ = \frac{1}{400} [x]_{300}^{500} \]
\[ = \frac{1}{400} (500 - 300) \]
\[ = \frac{1}{400} (200) \]
\[ = \frac{200}{400} = \frac{1}{2} \]
So, there is a 50% chance that daily sales will be between 300 and 500 litres. This makes sense for a uniform distribution since 300-500 is half of the total range (200-600).
In simple words: First, we find the constant 'k' by making sure the total chance over all possible milk sales adds up to one. Next, we write down the cumulative distribution function, which tells us the total chance of sales being less than or equal to any given amount 'x'. Finally, we use this 'k' value to find the chance that milk sales will be between 300 and 500 litres by integrating over that specific range.

🎯 Exam Tip: For uniform distributions, the probability for any sub-interval is proportional to its length. Always verify your integral calculation for \( P(a \le X \le b) \) by comparing it to the fraction of the total range that the sub-interval covers.

Question 4. The probability density function of X is given by
\[ f(x) = \begin{cases} ke^{-x/3} & \text{for } x > 0 \\ 0 & \text{for } x \le 0 \end{cases} \]
(i) the value of k
(ii) the distribution function
(iii) \( P(X < 3) \)
(iv) \( P(5 \le X) \)
(v) \( P(X \le 4) \)
Answer: This question involves an exponential probability distribution due to the \( e^{-x/3} \) term. We will use integration to find the constant \( k \) and then calculate the probabilities.
(i) To find the value of \( k \), the integral of the probability density function over its entire domain must equal 1.
\[ \int_{-\infty}^{\infty} f(x)dx = 1 \]
\[ \int_{-\infty}^{0} f(x)dx + \int_{0}^{\infty} f(x)dx = 1 \]
Since \( f(x) = 0 \) for \( x \le 0 \), the first integral is 0.
\[ 0 + \int_{0}^{\infty} ke^{-x/3}dx = 1 \]
\[ k \left[ \frac{e^{-x/3}}{-1/3} \right]_{0}^{\infty} = 1 \]
\[ k [-3e^{-x/3}]_{0}^{\infty} = 1 \]
\[ k (-3e^{-\infty} - (-3e^{0})) = 1 \]
\[ k (0 - (-3 \times 1)) = 1 \]
\[ k (3) = 1 \]
\[ k = \frac{1}{3} \]
(ii) The distribution function \( F(x) \) is \( F(x) = \int_{-\infty}^{x} f(u)du \).
**Case 1: \( x < 0 \)**
Since \( f(u) = 0 \) for \( u < 0 \), then \( F(x) = 0 \).
**Case 2: \( x \ge 0 \)**
Here we use \( k = \frac{1}{3} \) and integrate \( f(u) \) from 0 to \( x \).
\[ F(x) = \int_{-\infty}^{x} f(u)du = \int_{-\infty}^{0} f(u)du + \int_{0}^{x} f(u)du \]
\[ = 0 + \int_{0}^{x} \frac{1}{3}e^{-u/3}du \]
\[ = \frac{1}{3} \left[ \frac{e^{-u/3}}{-1/3} \right]_{0}^{x} \]
\[ = \frac{1}{3} [-3e^{-u/3}]_{0}^{x} \]
\[ = -e^{-x/3} - (-e^{0}) \]
\[ = -e^{-x/3} + 1 = 1 - e^{-x/3} \]
Combining both cases, the distribution function \( F(x) \) is:
\[ F(x) = \begin{cases} 0 & \text{for } x < 0 \\ 1 - e^{-x/3} & \text{for } x \ge 0 \end{cases} \]
(iii) To find \( P(X < 3) \), we integrate \( f(x) \) from 0 to 3 (since \( f(x) = 0 \) for negative values). Alternatively, we can use the distribution function \( F(x) \).
\[ P(X < 3) = \int_{0}^{3} f(x)dx = \int_{0}^{3} \frac{1}{3}e^{-x/3}dx \]
\[ = \frac{1}{3} \left[ \frac{e^{-x/3}}{-1/3} \right]_{0}^{3} \]
\[ = [-e^{-x/3}]_{0}^{3} \]
\[ = -e^{-3/3} - (-e^{0}) \]
\[ = -e^{-1} + 1 = 1 - e^{-1} \]
Using \( F(x) \): \( P(X < 3) = F(3) = 1 - e^{-3/3} = 1 - e^{-1} \).
(iv) To find \( P(5 \le X) \), we integrate \( f(x) \) from 5 to \( \infty \).
\[ P(5 \le X) = \int_{5}^{\infty} f(x)dx = \int_{5}^{\infty} \frac{1}{3}e^{-x/3}dx \]
\[ = \frac{1}{3} \left[ \frac{e^{-x/3}}{-1/3} \right]_{5}^{\infty} \]
\[ = [-e^{-x/3}]_{5}^{\infty} \]
\[ = -e^{-\infty} - (-e^{-5/3}) \]
\[ = 0 + e^{-5/3} = e^{-5/3} \]
Alternatively, using \( F(x) \): \( P(5 \le X) = 1 - P(X < 5) = 1 - F(5) = 1 - (1 - e^{-5/3}) = e^{-5/3} \).
(v) To find \( P(X \le 4) \), we use the distribution function \( F(x) \).
\[ P(X \le 4) = F(4) = 1 - e^{-4/3} \]
We integrate to confirm:
\[ P(X \le 4) = \int_{0}^{4} f(x)dx = \int_{0}^{4} \frac{1}{3}e^{-x/3}dx \]
\[ = \frac{1}{3} \left[ \frac{e^{-x/3}}{-1/3} \right]_{0}^{4} \]
\[ = [-e^{-x/3}]_{0}^{4} \]
\[ = -e^{-4/3} - (-e^{0}) \]
\[ = -e^{-4/3} + 1 = 1 - e^{-4/3} \]
The distribution function provides a quick way to calculate cumulative probabilities, while integration directly calculates the area under the curve for a specific range. Both methods yield consistent results.
In simple words: This problem is about an exponential probability, like how long something lasts. First, we find 'k' so that the total probability adds up to one. Then, we write down a special function, F(x), which tells us the chance that X is less than or equal to 'x'. Finally, we use either integration or this F(x) function to find the chances for X to be less than 3, greater than or equal to 5, and less than or equal to 4.

🎯 Exam Tip: For exponential distributions, remembering the formula for \( F(x) = 1 - e^{-\lambda x} \) (where \( \lambda = 1/3 \) in this case) can save time when calculating cumulative probabilities like \( P(X < a) \) or \( P(X \le a) \).

Question 5. If X is the random variable with probability density function f(x) given by,
\[ f(x) = \begin{cases} x+1 & \text{for } -1 \le x < 0 \\ -x+1 & \text{for } 0 \le x < 1 \\ 0 & \text{otherwise} \end{cases} \]
(i) the distribution function F(x)
(ii) \( P(-0.5 \le x \le 0.5) \)
Answer: This problem involves a piecewise-defined probability density function, meaning its rule changes at different intervals. We need to integrate to find the distribution function \( F(x) \) and specific probabilities.
(i) The distribution function \( F(x) \) is defined as \( F(x) = \int_{-\infty}^{x} f(u)du \). We need to consider different cases based on the intervals where \( f(x) \) changes.
**Case 1: \( x < -1 \)**
Since \( f(u) = 0 \) for \( u < -1 \), the integral is 0.
\[ F(x) = \int_{-\infty}^{x} f(u)du = 0 \]
**Case 2: \( -1 \le x < 0 \)**
Here, we integrate from -1 up to \( x \), using \( f(u) = u+1 \).
\[ F(x) = \int_{-\infty}^{x} f(u)du = \int_{-\infty}^{-1} f(u)du + \int_{-1}^{x} f(u)du \]
\[ = 0 + \int_{-1}^{x} (u+1)du \]
\[ = \left[ \frac{u^2}{2} + u \right]_{-1}^{x} \]
\[ = \left( \frac{x^2}{2} + x \right) - \left( \frac{(-1)^2}{2} + (-1) \right) \]
\[ = \left( \frac{x^2}{2} + x \right) - \left( \frac{1}{2} - 1 \right) \]
\[ = \frac{x^2}{2} + x - \left( -\frac{1}{2} \right) = \frac{x^2}{2} + x + \frac{1}{2} \]
**Case 3: \( 0 \le x < 1 \)**
We integrate through the previous intervals and then add the integral from 0 up to \( x \) using \( f(u) = -u+1 \).
\[ F(x) = \int_{-\infty}^{x} f(u)du = \int_{-\infty}^{-1} f(u)du + \int_{-1}^{0} f(u)du + \int_{0}^{x} f(u)du \]
\[ = 0 + \left[ \frac{u^2}{2} + u \right]_{-1}^{0} + \int_{0}^{x} (-u+1)du \]
First part: \( \left( \frac{0^2}{2} + 0 \right) - \left( \frac{(-1)^2}{2} + (-1) \right) = 0 - \left( \frac{1}{2} - 1 \right) = \frac{1}{2} \)
Second part: \( \left[ -\frac{u^2}{2} + u \right]_{0}^{x} = \left( -\frac{x^2}{2} + x \right) - \left( -\frac{0^2}{2} + 0 \right) = -\frac{x^2}{2} + x \)
So, \( F(x) = \frac{1}{2} + \left( -\frac{x^2}{2} + x \right) = -\frac{x^2}{2} + x + \frac{1}{2} \)
**Case 4: \( x \ge 1 \)**
At this point, all the probability has accumulated, so \( F(x) \) must be 1.
\[ F(x) = \int_{-\infty}^{x} f(u)du = \int_{-\infty}^{-1} f(u)du + \int_{-1}^{0} f(u)du + \int_{0}^{1} f(u)du + \int_{1}^{x} f(u)du \]
The first integral is 0. The second integral is \( \frac{1}{2} \). The third integral from 0 to 1 for \( (-u+1) \) is:
\[ \left[ -\frac{u^2}{2} + u \right]_{0}^{1} = \left( -\frac{1^2}{2} + 1 \right) - 0 = -\frac{1}{2} + 1 = \frac{1}{2} \]
The fourth integral from 1 to \( x \) is 0 because \( f(u) = 0 \) for \( u \ge 1 \).
So, \( F(x) = 0 + \frac{1}{2} + \frac{1}{2} + 0 = 1 \).
Combining all cases, the distribution function \( F(x) \) is:
\[ F(x) = \begin{cases} 0 & \text{for } x < -1 \\ \frac{x^2}{2} + x + \frac{1}{2} & \text{for } -1 \le x < 0 \\ -\frac{x^2}{2} + x + \frac{1}{2} & \text{for } 0 \le x < 1 \\ 1 & \text{for } x \ge 1 \end{cases} \]
(ii) To find \( P(-0.5 \le X \le 0.5) \), we can integrate \( f(x) \) over this range or use the distribution function \( F(x) \). Since the interval crosses \( x=0 \), we split the integral.
\[ P(-0.5 \le X \le 0.5) = \int_{-0.5}^{0.5} f(x)dx \]
\[ = \int_{-0.5}^{0} f(x)dx + \int_{0}^{0.5} f(x)dx \]
For \( -0.5 \le x < 0 \), \( f(x) = x+1 \). For \( 0 \le x \le 0.5 \), \( f(x) = -x+1 \).
\[ = \int_{-0.5}^{0} (x+1)dx + \int_{0}^{0.5} (-x+1)dx \]
\[ = \left[ \frac{x^2}{2} + x \right]_{-0.5}^{0} + \left[ -\frac{x^2}{2} + x \right]_{0}^{0.5} \]
First part: \( \left( \frac{0^2}{2} + 0 \right) - \left( \frac{(-0.5)^2}{2} + (-0.5) \right) = 0 - \left( \frac{0.25}{2} - 0.5 \right) = - (0.125 - 0.5) = 0.375 \)
Second part: \( \left( -\frac{(0.5)^2}{2} + 0.5 \right) - \left( -\frac{0^2}{2} + 0 \right) = \left( -\frac{0.25}{2} + 0.5 \right) - 0 = -0.125 + 0.5 = 0.375 \)
Adding them: \( P(-0.5 \le X \le 0.5) = 0.375 + 0.375 = 0.75 \)
Using \( F(x) \): \( P(-0.5 \le X \le 0.5) = F(0.5) - F(-0.5) \)
From Case 3 for \( F(x) \), for \( 0 \le x < 1 \): \( F(0.5) = -\frac{(0.5)^2}{2} + 0.5 + \frac{1}{2} = -\frac{0.25}{2} + 0.5 + 0.5 = -0.125 + 1 = 0.875 \)
From Case 2 for \( F(x) \), for \( -1 \le x < 0 \): \( F(-0.5) = \frac{(-0.5)^2}{2} + (-0.5) + \frac{1}{2} = \frac{0.25}{2} - 0.5 + 0.5 = 0.125 \)
So, \( P(-0.5 \le X \le 0.5) = 0.875 - 0.125 = 0.75 \)
Both methods yield the same result. The area under the curve between -0.5 and 0.5 is 0.75. This indicates a high probability for X to fall within this central range.
In simple words: First, we find the cumulative distribution function, F(x), which shows the total chance of X being up to a certain value. We have to do this in parts because the rule for the probability changes at different points. Then, we use this F(x) or direct integration to calculate the chance that X falls between -0.5 and 0.5.

🎯 Exam Tip: When dealing with piecewise functions, always carefully define the distribution function \( F(x) \) for each interval. This will simplify calculations for specific probabilities like \( P(a \le X \le b) \) as \( F(b) - F(a) \).

Question 6. If X is the random variable with distribution function F(x) given by,
\[ F(x) = \begin{cases} 0 & \text{for } x < 0 \\ \frac{1}{2}(x^2+x) & \text{for } 0 \le x < 1 \\ 1 & \text{for } 1 \le x < \infty \end{cases} \]
Find
(i) the probability density function f(x)
(ii) \( P(0.3 \le X \le 0.6) \)
Answer: This problem gives us the cumulative distribution function \( F(x) \) and asks for the probability density function \( f(x) \) and a specific probability. We know that \( f(x) \) is the derivative of \( F(x) \).
(i) To find the probability density function \( f(x) \), we differentiate \( F(x) \) with respect to \( x \) for each interval.
**For \( x < 0 \):**
\( F(x) = 0 \), so \( f(x) = F'(x) = 0 \)
**For \( 0 \le x < 1 \):**
\( F(x) = \frac{1}{2}(x^2+x) \)
\( f(x) = F'(x) = \frac{d}{dx} \left( \frac{1}{2}(x^2+x) \right) = \frac{1}{2}(2x+1) = x + \frac{1}{2} \)
**For \( 1 \le x < \infty \):**
\( F(x) = 1 \), so \( f(x) = F'(x) = 0 \)
Combining these results, the probability density function \( f(x) \) is:
\[ f(x) = \begin{cases} 0 & \text{for } x < 0 \\ x + \frac{1}{2} & \text{for } 0 \le x < 1 \\ 0 & \text{otherwise} \end{cases} \]
(ii) To find \( P(0.3 \le X \le 0.6) \), we can use the distribution function \( F(x) \), since \( P(a \le X \le b) = F(b) - F(a) \). Both 0.3 and 0.6 fall within the range \( 0 \le x < 1 \), where \( F(x) = \frac{1}{2}(x^2+x) \).
\[ P(0.3 \le X \le 0.6) = F(0.6) - F(0.3) \]
First, calculate \( F(0.6) \):
\[ F(0.6) = \frac{1}{2}((0.6)^2 + 0.6) = \frac{1}{2}(0.36 + 0.6) = \frac{1}{2}(0.96) = 0.48 \]
Next, calculate \( F(0.3) \):
\[ F(0.3) = \frac{1}{2}((0.3)^2 + 0.3) = \frac{1}{2}(0.09 + 0.3) = \frac{1}{2}(0.39) = 0.195 \]
Now, subtract \( F(0.3) \) from \( F(0.6) \):
\[ P(0.3 \le X \le 0.6) = 0.48 - 0.195 = 0.285 \]
The probability is 0.285. This means there is a 28.5% chance that X will fall within this specific range. The calculations show how the cumulative function helps find probabilities between two points without direct integration.
In simple words: First, we find the probability density function, f(x), by taking the derivative of the given distribution function, F(x). Then, to find the chance that X is between 0.3 and 0.6, we simply subtract the cumulative probability up to 0.3 from the cumulative probability up to 0.6.

🎯 Exam Tip: Always remember that the probability density function \( f(x) \) is the derivative of the distribution function \( F(x) \). This relationship is fundamental for converting between the two forms. When using \( F(b) - F(a) \) for \( P(a \le X \le b) \), ensure that both \( a \) and \( b \) fall within a continuous interval of \( F(x) \).