Let X and Y be two normal random variables, with means 0 and 3, respectively, and variances 1 and 16, respectively. Find the following, using the standard normal table. Express your answers to an accuracy of 3 decimal places.

Question

Standard Normal Table

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1 P(X > - 1) =

2. P(X <= - 2)

3. Let V = (4 - Y) / 3 Find the mean and the variance of V

[V] =

Var(V) =

P(- 2,Y <= 2) =

Answer 1

To solve the problems step by step, we need to standardize the normal random variables and then use the standard normal table.

1. \( P(X > -1) \)

Given that \( X \) is a normal random variable with mean \( \mu_X = 0 \) and variance \( \sigma_X^2 = 1 \), we can calculate the standard score (z-score):

\[ Z = \frac{X - \mu_X}{\sigma_X} = \frac{-1 - 0}{1} = -1 \]

Now, we want to find \( P(X > -1) \):

\[ P(X > -1) = 1 - P(X \leq -1) = 1 - P(Z \leq -1) \]

From the standard normal table, we find:

\[ P(Z \leq -1) \approx 0.1587 \]

Thus,

\[ P(X > -1) = 1 - 0.1587 = 0.8413 \]

2. \( P(X \leq -2) \)

Calculate the z-score for \( X = -2 \):

\[ Z = \frac{-2 - 0}{1} = -2 \]

Now, we find \( P(X \leq -2) \):

\[ P(X \leq -2) = P(Z \leq -2) \]

Using the standard normal table, we find:

\[ P(Z \leq -2) \approx 0.0228 \]

3. Let \( V = \frac{4 - Y}{3} \). Find the mean and variance of \( V \).

We are given \( Y \) is normal with mean \( \mu_Y = 3 \) and variance \( \sigma_Y^2 = 16 \).

Mean of \( V \):

Using the linear transformation property:

\[ \mu_V = \frac{4 - \mu_Y}{3} = \frac{4 - 3}{3} = \frac{1}{3} \approx 0.333 \]

Variance of \( V \):

Using the variance transformation formula:

\[ \text{Var}(V) = \left( \frac{1}{3} \right)^2 \text{Var}(Y) = \left( \frac{1}{3} \right)^2 \cdot 16 = \frac{16}{9} \approx 1.778 \]

4. \( P(-2 < Y \leq 2) \)

To find \( P(-2 < Y \leq 2) \), we first need to standardize both bounds using the mean and standard deviation of \( Y \).

For \( Y = -2 \):

\[ Z = \frac{-2 - 3}{4} = \frac{-5}{4} = -1.25 \]

\[ P(Y \leq -2) = P(Z \leq -1.25) \approx 0.1056 \]

For \( Y = 2 \):

\[ Z = \frac{2 - 3}{4} = \frac{-1}{4} = -0.25 \]

\[ P(Y \leq 2) = P(Z \leq -0.25) \approx 0.4013 \]

Now we can compute the desired probability:

\[ P(-2 < Y \leq 2) = P(Y \leq 2) - P(Y \leq -2) = 0.4013 - 0.1056 = 0.2957 \]

Summary of Results

1. \( P(X > -1) = 0.841 \)
2. \( P(X \leq -2) = 0.023 \)
3. \( \mu_V = 0.333 \), \( \text{Var}(V) = 1.778 \)
4. \( P(-2 < Y \leq 2) = 0.296 \)

Expressing to three decimal places, our results are:

• \( P(X > -1) = 0.841 \)
• \( P(X \leq -2) = 0.023 \)
• Mean of \( V \): \(0.333 \)
• Variance of \( V \): \( 1.778 \)
• \( P(-2 < Y \leq 2) = 0.296 \)