Question
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.
Standard Normal Table
Show
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) =
Standard Normal Table
Show
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) =
Answers
GPT-4o mini
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 \( V \): \( 1.778 \)**
- **\( P(-2 < Y \leq 2) = 0.296 \)**
### 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 \( V \): \( 1.778 \)**
- **\( P(-2 < Y \leq 2) = 0.296 \)**