Question

If \( Y = \log_{10} X \) has \( N(\mu, \sigma^2) \) distribution with moment generating function \( M_Y(t) = e^{\mu t + \frac{\sigma^2 t^2}{2}} \), then \( P(X<1000) \) equals

Hint:

When dealing with logarithmic transformations of normal variables, use the properties of the normal distribution to convert between scales and calculate probabilities using the standard normal distribution.

Answer 1

Correct Answer: 0.15 - 0.16

Step 1: Understanding the distribution.
Given that \( Y = \log_{10} X \) follows a normal distribution \( N(\mu, \sigma^2) \), we can express \( X \) as \( X = 10^Y \). The moment generating function of \( Y \) is given by \( M_Y(t) = e^{5t + 2t^2} \), which corresponds to a normal distribution with mean \( 5 \) and variance \( 2 \).

Step 2: Transforming the problem.
We need to find \( P(X<1000) \). Since \( X = 10^Y \), we can write the inequality \( X<1000 \) as \( Y<\log_{10} 1000 = 3 \). Thus, we need to find \( P(Y<3) \) for \( Y \sim N(5, 2) \).

Step 3: Calculating the probability.
To find \( P(Y<3) \), we standardize the variable \( Y \) by converting it to the standard normal distribution: \[ Z = \frac{Y - 5}{\sqrt{2}}. \] We then find the probability \( P(Z<\frac{3 - 5}{\sqrt{2}}) = P(Z<-\sqrt{2}) \). Using standard normal tables or a calculator, we find that \( P(Z<-\sqrt{2}) \) is approximately between 0.15 and 0.16.

Step 4: Conclusion.
Thus, the value of \( P(X<1000) \) is approximately \( (0.15, 0.16) \).