To find the standard deviation of a random variable $X$, we follow these steps:

1. Compute the mean ($\mu$) of $X$: $\mu = \sum_{x} x \cdot P(X = x)$

2. Compute the variance ($\sigma^2$): $\sigma^2 = \sum_{x} P(X = x) \cdot (x - \mu)^2$

3. Find the standard deviation ($\sigma$): $\sigma = \sqrt{\sigma^2}$

Step 1: Calculate $\mu$

Using the formula for the mean: $\mu = (1)(0.29) + (2)(0.22) + (3)(0.13) + (4)(0.16) + (5)(0.2)$ $\mu = 0.29 + 0.44 + 0.39 + 0.64 + 1.00 = 2.76$

Step 2: Calculate $\sigma^2$

Using the formula for variance: $\sigma^2 = \sum_{x} P(X = x) \cdot (x - \mu)^2$ First, calculate $(x - \mu)^2$ for each $x$:

• For $x = 1$: $(1 - 2.76)^2 = (-1.76)^2 = 3.0976$
• For $x = 2$: $(2 - 2.76)^2 = (-0.76)^2 = 0.5776$
• For $x = 3$: $(3 - 2.76)^2 = (0.24)^2 = 0.0576$
• For $x = 4$: $(4 - 2.76)^2 = (1.24)^2 = 1.5376$
• For $x = 5$: $(5 - 2.76)^2 = (2.24)^2 = 5.0176$

Now, compute $P(X = x) \cdot (x - \mu)^2$ for each $x$:

• For $x = 1$: $0.29 \cdot 3.0976 = 0.8983$
• For $x = 2$: $0.22 \cdot 0.5776 = 0.1271$
• For $x = 3$: $0.13 \cdot 0.0576 = 0.0075$
• For $x = 4$: $0.16 \cdot 1.5376 = 0.2460$
• For $x = 5$: $0.20 \cdot 5.0176 = 1.0035$

Summing these values: $\sigma^2 = 0.8983 + 0.1271 + 0.0075 + 0.2460 + 1.0035 = 2.2824$

Step 3: Calculate $\sigma$

$\sigma = \sqrt{\sigma^2} = \sqrt{2.2824} = 1.5114$

Final Answer:

The standard deviation of $X$ is: $\boxed{1.5114}$

Would you like to see a detailed explanation or verify with additional examples?

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5 Related Questions:

1. How is variance different from standard deviation in interpreting data?
2. What are other applications of the mean and standard deviation in statistics?
3. How do probability distributions relate to expected value and variance?
4. Can you explain the difference between discrete and continuous random variables?
5. How does changing the probabilities affect the mean and standard deviation?

Tip:

To minimize errors, double-check calculations involving $x - \mu$, as they are often a source of mistakes in variance computations.