Question
Compute the mean and standard deviation of the random variable with the given discrete probability distribution.
X| -8 -7 0 2 5
P(X)| .26 .1 .13 .27 .24
Find the mean, round to 3 decimal places if necessary
Find the standard deviation. Round to 3 decimal places if necessary.
X| -8 -7 0 2 5
P(X)| .26 .1 .13 .27 .24
Find the mean, round to 3 decimal places if necessary
Find the standard deviation. Round to 3 decimal places if necessary.
Answers
Answer
B is incorrect
Answer
what is the square root of 27.5443832
Answered by
GPT-4o mini
To compute the mean and standard deviation of the given discrete probability distribution, we will follow these steps:
### Step 1: Compute the Mean (Expected Value)
The mean \( \mu \) of a discrete random variable is calculated using the formula:
\[
\mu = \sum (X \cdot P(X))
\]
where \( X \) is the value of the random variable and \( P(X) \) is the probability of that value.
Given:
- \( X = [-8, -7, 0, 2, 5] \)
- \( P(X) = [0.26, 0.1, 0.13, 0.27, 0.24] \)
Now, we calculate the products \( X \cdot P(X) \):
\[
\begin{align*}
-8 \cdot 0.26 & = -2.08 \\
-7 \cdot 0.1 & = -0.7 \\
0 \cdot 0.13 & = 0 \\
2 \cdot 0.27 & = 0.54 \\
5 \cdot 0.24 & = 1.2 \\
\end{align*}
\]
Now we sum these products:
\[
\mu = -2.08 - 0.7 + 0 + 0.54 + 1.2 = -1.04
\]
### Step 2: Compute the Standard Deviation
The standard deviation \( \sigma \) is given by:
\[
\sigma = \sqrt{\sum [(X - \mu)^2 \cdot P(X)]}
\]
First, we calculate \( (X - \mu)^2 \):
\[
\begin{align*}
X = -8 & : (-8 - (-1.04))^2 = (-6.96)^2 = 48.4416 \\
X = -7 & : (-7 - (-1.04))^2 = (-5.96)^2 = 35.5216 \\
X = 0 & : (0 - (-1.04))^2 = (1.04)^2 = 1.0816 \\
X = 2 & : (2 - (-1.04))^2 = (3.04)^2 = 9.2416 \\
X = 5 & : (5 - (-1.04))^2 = (6.04)^2 = 36.4816 \\
\end{align*}
\]
Now we multiply by \( P(X) \):
\[
\begin{align*}
-8: & \ 48.4416 \cdot 0.26 = 12.598816 \\
-7: & \ 35.5216 \cdot 0.1 = 3.55216 \\
0: & \ 1.0816 \cdot 0.13 = 0.140608 \\
2: & \ 9.2416 \cdot 0.27 = 2.4972152 \\
5: & \ 36.4816 \cdot 0.24 = 8.755584 \\
\end{align*}
\]
Now, we sum these values:
\[
\sum [(X - \mu)^2 \cdot P(X)] = 12.598816 + 3.55216 + 0.140608 + 2.4972152 + 8.755584 = 27.5447832
\]
Now we take the square root to get the standard deviation:
\[
\sigma = \sqrt{27.5447832} \approx 5.25
\]
### Final Results
1. **Mean (rounded to 3 decimal places):**
\(\mu \approx -1.040\)
2. **Standard Deviation (rounded to 3 decimal places):**
\(\sigma \approx 5.250\)
Thus, the mean and standard deviation of the random variable are \( -1.040 \) and \( 5.250 \), respectively.
### Step 1: Compute the Mean (Expected Value)
The mean \( \mu \) of a discrete random variable is calculated using the formula:
\[
\mu = \sum (X \cdot P(X))
\]
where \( X \) is the value of the random variable and \( P(X) \) is the probability of that value.
Given:
- \( X = [-8, -7, 0, 2, 5] \)
- \( P(X) = [0.26, 0.1, 0.13, 0.27, 0.24] \)
Now, we calculate the products \( X \cdot P(X) \):
\[
\begin{align*}
-8 \cdot 0.26 & = -2.08 \\
-7 \cdot 0.1 & = -0.7 \\
0 \cdot 0.13 & = 0 \\
2 \cdot 0.27 & = 0.54 \\
5 \cdot 0.24 & = 1.2 \\
\end{align*}
\]
Now we sum these products:
\[
\mu = -2.08 - 0.7 + 0 + 0.54 + 1.2 = -1.04
\]
### Step 2: Compute the Standard Deviation
The standard deviation \( \sigma \) is given by:
\[
\sigma = \sqrt{\sum [(X - \mu)^2 \cdot P(X)]}
\]
First, we calculate \( (X - \mu)^2 \):
\[
\begin{align*}
X = -8 & : (-8 - (-1.04))^2 = (-6.96)^2 = 48.4416 \\
X = -7 & : (-7 - (-1.04))^2 = (-5.96)^2 = 35.5216 \\
X = 0 & : (0 - (-1.04))^2 = (1.04)^2 = 1.0816 \\
X = 2 & : (2 - (-1.04))^2 = (3.04)^2 = 9.2416 \\
X = 5 & : (5 - (-1.04))^2 = (6.04)^2 = 36.4816 \\
\end{align*}
\]
Now we multiply by \( P(X) \):
\[
\begin{align*}
-8: & \ 48.4416 \cdot 0.26 = 12.598816 \\
-7: & \ 35.5216 \cdot 0.1 = 3.55216 \\
0: & \ 1.0816 \cdot 0.13 = 0.140608 \\
2: & \ 9.2416 \cdot 0.27 = 2.4972152 \\
5: & \ 36.4816 \cdot 0.24 = 8.755584 \\
\end{align*}
\]
Now, we sum these values:
\[
\sum [(X - \mu)^2 \cdot P(X)] = 12.598816 + 3.55216 + 0.140608 + 2.4972152 + 8.755584 = 27.5447832
\]
Now we take the square root to get the standard deviation:
\[
\sigma = \sqrt{27.5447832} \approx 5.25
\]
### Final Results
1. **Mean (rounded to 3 decimal places):**
\(\mu \approx -1.040\)
2. **Standard Deviation (rounded to 3 decimal places):**
\(\sigma \approx 5.250\)
Thus, the mean and standard deviation of the random variable are \( -1.040 \) and \( 5.250 \), respectively.
Answered by
GPT-4o mini
Let's recalculate the standard deviation step-by-step to ensure accuracy.
### Step 1: Mean Calculation (Reiteration)
First, let's reiterate how we calculated the mean:
The mean (expected value) \(\mu\) is computed as follows:
\[
\mu = \sum (X \cdot P(X))
\]
where:
- \(X = [-8, -7, 0, 2, 5]\)
- \(P(X) = [0.26, 0.1, 0.13, 0.27, 0.24]\)
Calculating each term:
\[
\begin{align*}
-8 \cdot 0.26 & = -2.08 \\
-7 \cdot 0.1 & = -0.7 \\
0 \cdot 0.13 & = 0 \\
2 \cdot 0.27 & = 0.54 \\
5 \cdot 0.24 & = 1.2 \\
\end{align*}
\]
Now, summing these products:
\[
\mu = -2.08 - 0.7 + 0 + 0.54 + 1.2 = -1.04
\]
### Step 2: Standard Deviation Calculation
The standard deviation \( \sigma \) is given by:
\[
\sigma = \sqrt{\sum [(X - \mu)^2 \cdot P(X)]}
\]
1. **Calculate \( (X - \mu)^2 \)** using \( \mu = -1.04 \):
\[
(X - \mu)^2:
\begin{align*}
X = -8 & : (-8 + 1.04)^2 = (-6.96)^2 = 48.4416\\
X = -7 & : (-7 + 1.04)^2 = (-5.96)^2 = 35.5216\\
X = 0 & : (0 + 1.04)^2 = (1.04)^2 = 1.0816\\
X = 2 & : (2 + 1.04)^2 = (3.04)^2 = 9.2416\\
X = 5 & : (5 + 1.04)^2 = (6.04)^2 = 36.4816\\
\end{align*}
\]
2. **Multiply by \( P(X) \)**:
Now multiply the squared differences by the corresponding probabilities:
\[
\begin{align*}
-8: & \ 48.4416 \cdot 0.26 = 12.598816\\
-7: & \ 35.5216 \cdot 0.1 = 3.55216\\
0: & \ 1.0816 \cdot 0.13 = 0.140608\\
2: & \ 9.2416 \cdot 0.27 = 2.4972152\\
5: & \ 36.4816 \cdot 0.24 = 8.755584\\
\end{align*}
\]
3. **Sum these values**:
\[
\begin{align*}
\text{Total} & = 12.598816 + 3.55216 + 0.140608 + 2.4972152 + 8.755584\\
& = 27.5447832
\end{align*}
\]
4. **Calculate the standard deviation**:
Now, take the square root of the total:
\[
\sigma = \sqrt{27.5447832}
\]
Calculating this gives:
\[
\sigma \approx 5.25
\]
### Double-check Calculation of the Standard Deviation
1. Calculate the contributions from each piece specifically:
\[
\begin{align*}
\text{For } -8: & \ 48.4416 \cdot 0.26 = 12.598816\\
\text{For } -7: & \ 35.5216 \cdot 0.1 = 3.55216\\
\text{For } 0: & \ 1.0816 \cdot 0.13 = 0.140608\\
\text{For } 2: & \ 9.2416 \cdot 0.27 = 2.4972152\\
\text{For } 5: & \ 36.4816 \cdot 0.24 = 8.755584\\
\end{align*}
\]
Re-summing these correctly:
\[
12.598816 + 3.55216 + 0.140608 + 2.4972152 + 8.755584 = 27.5447832
\]
Finally, calculating:
\[
\sigma \approx \sqrt{27.5447832} \approx 5.25
\]
### Conclusion
The calculations suggest that the **mean** is \( -1.040 \) and the **standard deviation** is \( 5.250 \). If these values are still incorrect, could you please provide the expected results or specify what part might be erroneous?
### Step 1: Mean Calculation (Reiteration)
First, let's reiterate how we calculated the mean:
The mean (expected value) \(\mu\) is computed as follows:
\[
\mu = \sum (X \cdot P(X))
\]
where:
- \(X = [-8, -7, 0, 2, 5]\)
- \(P(X) = [0.26, 0.1, 0.13, 0.27, 0.24]\)
Calculating each term:
\[
\begin{align*}
-8 \cdot 0.26 & = -2.08 \\
-7 \cdot 0.1 & = -0.7 \\
0 \cdot 0.13 & = 0 \\
2 \cdot 0.27 & = 0.54 \\
5 \cdot 0.24 & = 1.2 \\
\end{align*}
\]
Now, summing these products:
\[
\mu = -2.08 - 0.7 + 0 + 0.54 + 1.2 = -1.04
\]
### Step 2: Standard Deviation Calculation
The standard deviation \( \sigma \) is given by:
\[
\sigma = \sqrt{\sum [(X - \mu)^2 \cdot P(X)]}
\]
1. **Calculate \( (X - \mu)^2 \)** using \( \mu = -1.04 \):
\[
(X - \mu)^2:
\begin{align*}
X = -8 & : (-8 + 1.04)^2 = (-6.96)^2 = 48.4416\\
X = -7 & : (-7 + 1.04)^2 = (-5.96)^2 = 35.5216\\
X = 0 & : (0 + 1.04)^2 = (1.04)^2 = 1.0816\\
X = 2 & : (2 + 1.04)^2 = (3.04)^2 = 9.2416\\
X = 5 & : (5 + 1.04)^2 = (6.04)^2 = 36.4816\\
\end{align*}
\]
2. **Multiply by \( P(X) \)**:
Now multiply the squared differences by the corresponding probabilities:
\[
\begin{align*}
-8: & \ 48.4416 \cdot 0.26 = 12.598816\\
-7: & \ 35.5216 \cdot 0.1 = 3.55216\\
0: & \ 1.0816 \cdot 0.13 = 0.140608\\
2: & \ 9.2416 \cdot 0.27 = 2.4972152\\
5: & \ 36.4816 \cdot 0.24 = 8.755584\\
\end{align*}
\]
3. **Sum these values**:
\[
\begin{align*}
\text{Total} & = 12.598816 + 3.55216 + 0.140608 + 2.4972152 + 8.755584\\
& = 27.5447832
\end{align*}
\]
4. **Calculate the standard deviation**:
Now, take the square root of the total:
\[
\sigma = \sqrt{27.5447832}
\]
Calculating this gives:
\[
\sigma \approx 5.25
\]
### Double-check Calculation of the Standard Deviation
1. Calculate the contributions from each piece specifically:
\[
\begin{align*}
\text{For } -8: & \ 48.4416 \cdot 0.26 = 12.598816\\
\text{For } -7: & \ 35.5216 \cdot 0.1 = 3.55216\\
\text{For } 0: & \ 1.0816 \cdot 0.13 = 0.140608\\
\text{For } 2: & \ 9.2416 \cdot 0.27 = 2.4972152\\
\text{For } 5: & \ 36.4816 \cdot 0.24 = 8.755584\\
\end{align*}
\]
Re-summing these correctly:
\[
12.598816 + 3.55216 + 0.140608 + 2.4972152 + 8.755584 = 27.5447832
\]
Finally, calculating:
\[
\sigma \approx \sqrt{27.5447832} \approx 5.25
\]
### Conclusion
The calculations suggest that the **mean** is \( -1.040 \) and the **standard deviation** is \( 5.250 \). If these values are still incorrect, could you please provide the expected results or specify what part might be erroneous?
Answered by
GPT-4o mini
The square root of \(27.5443832\) is approximately:
\[
\sqrt{27.5443832} \approx 5.25
\]
To be more precise, it is approximately \(5.25\) when rounded to two decimal places, or more specifically:
\[
\sqrt{27.5443832} \approx 5.252
\]
So if you round to three decimal places, you would get:
\[
\sqrt{27.5443832} \approx 5.252
\]
Thus, the standard deviation \(\sigma\) should be reported as approximately **5.252** when rounded to three decimal places. Thank you for your patience!
\[
\sqrt{27.5443832} \approx 5.25
\]
To be more precise, it is approximately \(5.25\) when rounded to two decimal places, or more specifically:
\[
\sqrt{27.5443832} \approx 5.252
\]
So if you round to three decimal places, you would get:
\[
\sqrt{27.5443832} \approx 5.252
\]
Thus, the standard deviation \(\sigma\) should be reported as approximately **5.252** when rounded to three decimal places. Thank you for your patience!