Question

X : P(x):

0 0.05
1 0.25
2 0.15
3 0.55 Find the standard deviation of this probability distribution. Give your answer to at least 2 decimal places

Answer 1

To find the standard deviation of the probability distribution, we first need to calculate the mean of the distribution, which is given by:

μ = Σ(x * P(x))

where x is the value of the random variable and P(x) is the corresponding probability.

Using the values given in the problem, we can calculate the mean as:

μ = (0 * 0.05) + (1 * 0.25) + (2 * 0.15) + (3 * 0.55) = 2.4

Next, we need to calculate the variance of the distribution, which is given by:

σ^2 = Σ[(x - μ)^2 * P(x)]

Using the values given in the problem, we can calculate the variance as:

σ^2 = [(0 - 2.4)^2 * 0.05] + [(1 - 2.4)^2 * 0.25] + [(2 - 2.4)^2 * 0.15] + [(3 - 2.4)^2 * 0.55] ≈ 0.73

Finally, we can calculate the standard deviation as the square root of the variance:

σ = sqrt(σ^2) ≈ 0.85

Therefore, the standard deviation of the probability distribution is approximately 0.85, rounded to two decimal places.

Answer 2

the standard deviation of this probability distribution is approximately 1.39.

Explanation:

To find the standard deviation of a probability distribution, we use the formula:

σ = sqrt[Σ(x - μ)^2P(x)]

where σ is the standard deviation, x is the value of the random variable, P(x) is the probability of that value, μ is the mean of the distribution.

First, we need to find the mean of the distribution:

μ = ΣxP(x)

= 0(0.05) + 1(0.25) + 2(0.15) + 3(0.55)

= 1.9

Next, we can calculate the standard deviation:

σ = sqrt[Σ(x - μ)^2P(x)]

= sqrt[(0 - 1.9)^2(0.05) + (1 - 1.9)^2(0.25) + (2 - 1.9)^2(0.15) + (3 - 1.9)^2(0.55)]

= sqrt[0.9025 + 0.1225 + 0.0225 + 0.8925]

= sqrt(1.94)

≈ 1.39

Therefore, the standard deviation of this probability distribution is approximately 1.39.