Question

Find the standard deviation of the following data. Round your answer to one decimal place. x 0 1 2 3 4 5 P(X

Answer 1

The standard deviation of the data is 1.8

X ____ 0 __ 1 __ 2 __ 3 __ 4 __ 5

P(X) _ 0.2 _ 0.1 _ 0.1 _ 0.2_ 0.2 _ 0.2

The standard deviation is defined as :

• Std(X) = √Var(X)

Var(X) = ∑x² *p(x) - E(X)²

E(X) = ∑x*p(x)

E(X) = (0*0.2) + (1*0.1) + (2*0.1) + (3*0.2) + (4*0.2) + (5*0.2)

E(X) = 2.7

∑x² * p(x) = (0²*0.2) + (1²*0.1) + (2²*0.1) + (3²*0.2) + (4²*0.2) + (5²*0.2) = 10.5

Var(X) = 10.5 - 2.7²

Var(X) = 3.21

Std(X) = √3.21

Std(X) = 1.79 ≈ 1.8

Complete Question:

Find the standard deviation of the following data. Round your answer to one decimal place. x 0 1 2 3 4 5 P(X) 0.2 0.1 0.1 0.2 0.2 0.2

Answer 2

`\sigma = 1.8`

Explanation:

Given

`\begin{array}{ccccccc}x & {0} & {1} & {2} & {3} & {4}& {5} \ \\ P(x) & {0.2} & {0.1} & {0.1} & {0.2} & {0.2}& {0.2} \ \end{array}`

Required

The standard deviation

First, calculate the expected value E(x)

`E(x) = \sum x * P(x)`

So, we have:

`E(x) = 0 * 0.2 + 1 * 0.1 + 2 * 0.1 + 3 * 0.2 + 4 * 0.2 + 5 * 0.2`

`E(x) = 2.7`

Next, calculate E(x^2)

`E(x^2) = \sum x^2 * P(x)`

So, we have:

`E(x^2) = 0^2 * 0.2 + 1^2 * 0.1 + 2^2 * 0.1 + 3^2 * 0.2 + 4^2 * 0.2 + 5^2 * 0.2`

`E(x^2) = 10.5`

The standard deviation is:

`\sigma = √(E(x^2) - (E(x))^2)`

`\sigma = √(10.5 - 2.7^2)`

`\sigma = √(10.5 - 7.29)`

`\sigma = √(3.21)`

`\sigma = 1.8` --- approximated