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

x| -5, 1, 4, 6, 8
P(x)| 0.11, 0.26, 0.23, 0.28, 0.12
(a) Find the mean. Round the answer to three decimal places, if necessary.
(b) Find the standard deviation. Round the answer to three decimal places, if necessary.

7 answers

To find the mean and standard deviation of the discrete random variable given its probability distribution, we can use the following formulas:

Mean (µ):

The mean of a discrete probability distribution is calculated using the formula:

\[ \mu = \sum (x \cdot P(x)) \]

Standard Deviation (σ):

The standard deviation is calculated using the formula:

\[ \sigma = \sqrt{\sum (x^2 \cdot P(x)) - \mu^2} \]

Given Data:

Let's summarize the provided data:

\[ \begin{array}{|c|c|} \hline x & P(x) \ \hline -5 & 0.11 \ 1 & 0.26 \ 4 & 0.23 \ 6 & 0.28 \ 8 & 0.12 \ \hline \end{array} \]

(a) Calculate the Mean (µ):

First, we compute \(x \cdot P(x)\) for each value:

  1. \(x = -5\): \(-5 \cdot 0.11 = -0.55\)
  2. \(x = 1\): \(1 \cdot 0.26 = 0.26\)
  3. \(x = 4\): \(4 \cdot 0.23 = 0.92\)
  4. \(x = 6\): \(6 \cdot 0.28 = 1.68\)
  5. \(x = 8\): \(8 \cdot 0.12 = 0.96\)

Now sum these products:

\[ \mu = -0.55 + 0.26 + 0.92 + 1.68 + 0.96 = 2.27 \]

(b) Calculate the Standard Deviation (σ):

Next, we compute \(x^2 \cdot P(x)\):

  1. \(x = -5\): \((-5)^2 \cdot 0.11 = 25 \cdot 0.11 = 2.75\)
  2. \(x = 1\): \(1^2 \cdot 0.26 = 1 \cdot 0.26 = 0.26\)
  3. \(x = 4\): \(4^2 \cdot 0.23 = 16 \cdot 0.23 = 3.68\)
  4. \(x = 6\): \(6^2 \cdot 0.28 = 36 \cdot 0.28 = 10.08\)
  5. \(x = 8\): \(8^2 \cdot 0.12 = 64 \cdot 0.12 = 7.68\)

Now sum these products:

\[ \sum (x^2 \cdot P(x)) = 2.75 + 0.26 + 3.68 + 10.08 + 7.68 = 24.45 \]

Now we can compute the variance:

\[ \sigma^2 = \sum (x^2 \cdot P(x)) - \mu^2 = 24.45 - (2.27)^2 \] \[ (2.27)^2 = 5.1529 \] \[ \sigma^2 = 24.45 - 5.1529 = 19.2971 \]

Now take the square root to find the standard deviation:

\[ \sigma = \sqrt{19.2971} \approx 4.39 \]

Final Answers:

(a) Mean (µ): 2.270

(b) Standard Deviation (σ): 4.390

Both answers are incorrect..

Let's go through the calculations step-by-step to ensure we get the correct results for the mean and standard deviation of the given discrete probability distribution.

Given Data:

\[ \begin{array}{|c|c|} \hline x & P(x) \ \hline -5 & 0.11 \ 1 & 0.26 \ 4 & 0.23 \ 6 & 0.28 \ 8 & 0.12 \ \hline \end{array} \]

(a) Calculate the Mean (µ):

Using the formula for the mean:

\[ \mu = \sum (x \cdot P(x)) \]

We will compute each \( x \cdot P(x) \):

  1. For \( x = -5 \): \[ -5 \cdot 0.11 = -0.55 \]

  2. For \( x = 1 \): \[ 1 \cdot 0.26 = 0.26 \]

  3. For \( x = 4 \): \[ 4 \cdot 0.23 = 0.92 \]

  4. For \( x = 6 \): \[ 6 \cdot 0.28 = 1.68 \]

  5. For \( x = 8 \): \[ 8 \cdot 0.12 = 0.96 \]

Now, we sum these results:

\[ \mu = -0.55 + 0.26 + 0.92 + 1.68 + 0.96 = 2.27 \]

Thus, the mean \(\mu\) is:

\[ \mu = 2.27 \]

(b) Calculate the Standard Deviation (σ):

To calculate the standard deviation, we first need to compute \( x^2 \cdot P(x) \):

  1. For \( x = -5 \): \[ (-5)^2 \cdot 0.11 = 25 \cdot 0.11 = 2.75 \]

  2. For \( x = 1 \): \[ 1^2 \cdot 0.26 = 1 \cdot 0.26 = 0.26 \]

  3. For \( x = 4 \): \[ 4^2 \cdot 0.23 = 16 \cdot 0.23 = 3.68 \]

  4. For \( x = 6 \): \[ 6^2 \cdot 0.28 = 36 \cdot 0.28 = 10.08 \]

  5. For \( x = 8 \): \[ 8^2 \cdot 0.12 = 64 \cdot 0.12 = 7.68 \]

Now sum these products:

\[ \sum (x^2 \cdot P(x)) = 2.75 + 0.26 + 3.68 + 10.08 + 7.68 = 24.45 \]

Calculate the variance using:

\[ \sigma^2 = \sum (x^2 \cdot P(x)) - \mu^2 \]

Calculating \( \mu^2 \):

\[ \mu^2 = (2.27)^2 = 5.1529 \]

Now calculate variance:

\[ \sigma^2 = 24.45 - 5.1529 = 19.2971 \]

Finally, taking the square root gives us the standard deviation:

\[ \sigma = \sqrt{19.2971} \approx 4.39 \]

Correct Final Answers

  1. (a) Mean (µ): 2.270
  2. (b) Standard Deviation (σ): 4.390

If you could provide the expected values, we can also cross-check if any assumptions of the input data were misinterpreted.

what is the square root of 15.68

The square root of 15.68 is approximately:

\[ \sqrt{15.68} \approx 3.96 \]

If you need a more precise value, it's:

\[ \sqrt{15.68} \approx 3.96 \]

So the value rounded to two decimal places is 3.96.

As answer is the mean is 3.27.

Now find the standard deviation for B

Let's first summarize the calculations again based on the corrected mean of \( \mu = 3.27 \). We will compute the standard deviation using the provided probabilities.

Given Data:

\[ \begin{array}{|c|c|} \hline x & P(x) \ \hline -5 & 0.11 \ 1 & 0.26 \ 4 & 0.23 \ 6 & 0.28 \ 8 & 0.12 \ \hline \end{array} \]

Calculate Standard Deviation (σ):

We use the formula for standard deviation:

\[ \sigma = \sqrt{\sum (x^2 \cdot P(x)) - \mu^2} \]

Step 1: Calculate \( x^2 \cdot P(x) \)

  1. For \( x = -5 \): \[ (-5)^2 \cdot 0.11 = 25 \cdot 0.11 = 2.75 \]

  2. For \( x = 1 \): \[ 1^2 \cdot 0.26 = 1 \cdot 0.26 = 0.26 \]

  3. For \( x = 4 \): \[ 4^2 \cdot 0.23 = 16 \cdot 0.23 = 3.68 \]

  4. For \( x = 6 \): \[ 6^2 \cdot 0.28 = 36 \cdot 0.28 = 10.08 \]

  5. For \( x = 8 \): \[ 8^2 \cdot 0.12 = 64 \cdot 0.12 = 7.68 \]

Step 2: Sum these products

\[ \sum (x^2 \cdot P(x)) = 2.75 + 0.26 + 3.68 + 10.08 + 7.68 = 24.45 \]

Step 3: Calculate \( \mu^2 \)

Given \( \mu = 3.27 \): \[ \mu^2 = (3.27)^2 = 10.7129 \]

Step 4: Calculate Variance

\[ \sigma^2 = \sum (x^2 \cdot P(x)) - \mu^2 = 24.45 - 10.7129 = 13.7371 \]

Step 5: Calculate Standard Deviation

Now take the square root to find the standard deviation: \[ \sigma = \sqrt{13.7371} \approx 3.71 \]

Final Answer

The standard deviation (σ) is approximately 3.710 when rounded to three decimal places.