Question

A sociologist randomly selects single adults for different groups of​ three, and the random variable x is the number in the group who say that the most fun way to flirt is in person. Determine whether a probability distribution is given. If a probability distribution is​ given, find its mean and standard deviation. If a probability distribution is not​ given, identify the requirements that are not satisfied. x ​P(x) 0 0.087 1 0.344 2 0.415 3 0.154

Answer 1

Yes the table shows a probability distribution

The mean is
`E(X) = 1.636`

The standard deviation is
`\sigma = 0.845`

Explanation:

From the question we are told that

The data given is

x 0 1 2 3

P(x) 0.087 0.344 0.415 0.154

Generally we can evaluate
`P(x) * x` as

`P(x) * x` 0 (1 * 0.344 ) (2 * 0.415) (3 * 0.154 )

=>
`P(x) * x` 0 0.344 0.830 0.462

Generally we can evaluate
`P(x) * x^2` as

`P(x) * x^2` 0 [tex ](1^2 * 0.344 )[/tex] [tex ](2^2 * 0.415)[/tex] [tex ](3^2 * 0.154 )[/tex]

=>
`P(x) * x^2` 0
`0.344` [tex ]1.66 [/tex]
`1.386`

Generally the mean is mathematically represented as

`E(X) = \sum [x* P(x)]`

`E(X) = 0 + 0.344+ 0.830+0.462`

`E(X) = 1.636`

Generally the standard deviation is mathematically represented as

`\sigma = √(V(X))`

Here
`V(X)` is the variance which is mathematically represented as

`V(X) = [E(X^2) ]- [E(X)]^2`

Here

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

`E(X) =0 +0.344+1.66+1.386`

`E(X) = 3.3900`

So

`V(X) = 3.3900 - [1.636]^2`

`V(X) = 3.3900 - 2.6764`

`V(X) = 0.7136`

So

`\sigma = √(0.7136)`

`\sigma = 0.845`