Question

If the error involved in making a certain measurement is continuous random variable \( X \) with probability density function \( f(x) = k (4 - x^2) \) for \( -2 \leq x \leq 2 \), and \( f(x) = 0 \) otherwise, then \[ P(|-1<X<1|) \]

• A. \( \frac{13}{16} \)
• B. \( \frac{1}{2} \)
• C. \( \frac{1}{3} \)
• D. \( \frac{11}{16} \)

Hint:

For continuous random variables, always normalize the probability density function before calculating the probabilities.

Answer 1

The Correct Option is D

Step 1: Find the value of \( k \).
To find \( k \), we use the fact that the total probability must equal 1. The probability density function \( f(x) \) is given by: \[ \int_{-2}^{2} k(4 - x^2) dx = 1 \] Evaluating the integral: \[ k \left[ 4x - \frac{x^3}{3} \right]_{-2}^{2} = 1 \] Substitute the limits: \[ k \left[ (4(2) - \frac{2^3}{3}) - (4(-2) - \frac{(-2)^3}{3}) \right] = 1 \] Simplifying: \[ k \left[ (8 - \frac{8}{3}) - (-8 + \frac{8}{3}) \right] = 1 \] \[ k \left[ 8 - \frac{8}{3} + 8 - \frac{8}{3} \right] = 1 \] \[ k \left[ 16 - \frac{16}{3} \right] = 1 \] \[ k \left[ \frac{48}{3} - \frac{16}{3} \right] = 1 \] \[ k \left[ \frac{32}{3} \right] = 1 \] Thus, \( k = \frac{3}{32} \). Step 2: Find the probability.
Now, to find \( P(-1<X<1) \), we integrate \( f(x) \) from \( x = -1 \) to \( x = 1 \): \[ P(-1<X<1) = \int_{-1}^{1} \frac{3}{32} (4 - x^2) dx \] This simplifies to: \[ P(-1<X<1) = \frac{3}{32} \int_{-1}^{1} (4 - x^2) dx \] Evaluating the integral: \[ \frac{3}{32} \left[ 4x - \frac{x^3}{3} \right]_{-1}^{1} \] Substituting the limits: \[ \frac{3}{32} \left[ (4(1) - \frac{1^3}{3}) - (4(-1) - \frac{(-1)^3}{3}) \right] \] Simplifying: \[ \frac{3}{32} \left[ (4 - \frac{1}{3}) - (-4 + \frac{1}{3}) \right] \] \[ \frac{3}{32} \left[ 4 - \frac{1}{3} + 4 - \frac{1}{3} \right] \] \[ \frac{3}{32} \left[ 8 - \frac{2}{3} \right] \] \[ \frac{3}{32} \times \frac{22}{3} = \frac{11}{16} \] Step 3: Conclusion.
Thus, the probability is \( \boxed{\frac{11}{16}} \).