Question

Consider the normal probability distribution function \[ f(x) = \frac{4}{\sqrt{2\pi}} e^{-8(x+3)^2}. \] If \( \mu \) and \( \sigma \) are the mean and standard deviation of \( f(x) \) respectively, then the ordered pair \( (\mu, \sigma) \) is:

• A. \( \left( 3, \frac{1}{4} \right) \)
• B. \( \left( -3, \frac{1}{4} \right) \)
• C. \( \left( 3, 4 \right) \)
• D. \( \left( -3, 4 \right) \)

Hint:

In normal distribution problems, identify the mean (\( \mu \)) from the shift in \( x \), and calculate the standard deviation (\( \sigma \)) using the coefficient of \( x^2 \) in the exponential.

Answer 1

The Correct Option is B

Step 1: Analyze the general form of the normal distribution. The general form of the normal probability distribution is: \[ f(x) = \frac{1}{\sqrt{2\pi} \sigma} e^{-\frac{(x - \mu)^2}{2\sigma^2}}, \] where \( \mu \) is the mean, and \( \sigma \) is the standard deviation. Step 2: Match the given equation to the general form. The given function is: \[ f(x) = \frac{4}{\sqrt{2\pi}} e^{-8(x+3)^2}. \] Compare this with the general form. The coefficient of \( (x + 3)^2 \) inside the exponent is \( 8 \). This gives: \[ \frac{1}{2\sigma^2} = 8 \quad \Rightarrow \quad \sigma^2 = \frac{1}{16} \quad \Rightarrow \quad \sigma = \frac{1}{4}. \] The mean \( \mu \) is obtained from the shift \( (x + 3) \), which indicates \( \mu = -3 \). Step 3: Conclusion. The ordered pair \( (\mu, \sigma) \) is \( \left( -3, \frac{1}{4} \right) \).