Question

For the following listed, the x value is first, the px is second... 0, .236, 1, .396, 2, .264, 3, .088, 4, .015, 5, .001. Need to find the mean, expected number, and standard deviation.

Answer 1

Conditions

For the following listed, the x value is first, the px is second... 0, .236, 1, .396, 2, .264, 3, .088, 4, .015, 5, .001. Need to find the mean, expected number, and standard deviation.

Solution

An expected value (or mean) could be found by using following formula:

$$M(x) = \sum_{i=1}^{6} x_i P(x_i) = 0 \cdot 0.236 + 1 \cdot 0.396 + 2 \cdot 0.264 + 3 \cdot 0.088 + 4 \cdot 0.015 + 5 \cdot 0.001 = 1.253$$

A standard deviation could be found by using following formula:

$$\sigma = \sqrt{M(x^2) - (M(x))^2}$$

Let's calculate $x^2$ values and its probabilities:

$$M(x^2) = \sum_{i=1}^{6} x_i^2 P(x_i^2) = 0 \cdot 0.236 + 1 \cdot 0.396 + 4 \cdot 0.264 + 9 \cdot 0.088 + 16 \cdot 0.015 + 25 \cdot 0.001 = 2.509$$ $$\sigma = \sqrt{M(x^2) - (M(x))^2} = \sqrt{2.509 - 1.253^2} = \sqrt{0.938991} = 0.969015$$

Answer:

$$M(x) = 1.253$$ $$\sigma = 0.969015$$