Compute the mean and standard deviation of the random variable with the given discrete probability distribution.

Question

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.

Answer 1

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.

Answer 2

B is incorrect

Answer 3

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?

Answer 4

what is the square root of 27.5443832

Answer 5

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!