Math / Data & Statistics

Questionc. Compute the variance and standard deviation of the probability distribution. Round each result to 4 decimal places, if necessary. \begin{tabular}{|r|r|r|r|r|r|} \hline $x$ & $P(x)$ & $x \cdot P(x)$ & $x-\mu$ & $(x-\mu)^{2}$ & $(x-\mu)^{2} \cdot P(x)$ \\ \hline 3 & 0.19 & 0.57 & -1.66 & 2.7556 & 0.5236 \\ \hline 4 & 0.2 & 0.8 & -0.66 & 0.4356 & 0.0871 \\ \hline 5 & 0.37 & 1.85 & 0.34 & 0.1156 & 0.0428 \\ \hline 6 & 0.24 & 1.44 & 1.34 & 1.7956 & 0.4309 \\ \hline \end{tabular} $\begin{array}{l} \sigma_{X}^{2}=\square \\ \sigma_{X}=\square \end{array}$ d. What is the probability that $x \leq 4$ ? $\square$

Studdy Solution

STEP 1

1. The probability distribution is discrete.
2. The variance is calculated using the formula $\sigma^2 = \sum (x - \mu)^2 \cdot P(x)$ .
3. The standard deviation is the square root of the variance.
4. The probability that $x \leq 4$ is the sum of probabilities for $x = 3$ and $x = 4$ .

STEP 2

1. Calculate the variance.
2. Calculate the standard deviation.
3. Calculate the probability that $x \leq 4$ .

STEP 3

Calculate the variance using the provided values in the table:
$\sigma^2 = \sum (x - \mu)^2 \cdot P(x) = 0.5236 + 0.0871 + 0.0428 + 0.4309$
$\sigma^2 = 1.0844$

STEP 4

Calculate the standard deviation by taking the square root of the variance:
$\sigma = \sqrt{1.0844} \approx 1.0413$

STEP 5

Calculate the probability that $x \leq 4$ :
$P(x \leq 4) = P(x = 3) + P(x = 4) = 0.19 + 0.2 = 0.39$
The variance is:
$\sigma^2 = \boxed{1.0844}$
The standard deviation is:
$\sigma = \boxed{1.0413}$
The probability that $x \leq 4$ is:
$\boxed{0.39}$

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