Question

Fifty-three percent of employees make judgements about their co-workers based on the cleanliness of their desk. You randomly select 8 employees and ask them if they judge co-workers based on this criterion. The random variable is the number of employees who judge their co-workers by cleanliness. Which outcomes of this binomial distribution would be considered unusual?

Answer 1

No employees or 8 employees judging co-workers based on the cleanliness of their desk would be considered unusual outcomes.

Explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

The expected value of the binomial distribution is:

`E(X) = np`

The standard deviation of the binomial distribution is:

`√(V(X)) = √(np(1-p))`

Outcomes are unusual when they are more than 2.5 standard deviations of the mean.

In this problem, we have that:

`n = 8, p = 0.53`

So

`E(X) = np = 8*0.53 = 4.24`

`√(V(X)) = √(np(1-p)) = 1.41`

`4.24 - 2.5*1.41 = 0.71`

`4.24 + 2.5*1.41 = 7.77`

No employees or 8 employees judging co-workers based on the cleanliness of their desk would be considered unusual outcomes.

Answer 2

The unusual
`X` values ​​for this model are:
`X = 0, 1, 2, 7, 8`

Explanation:

A binomial random variable
`X` represents the number of successes obtained in a repetition of
`n` Bernoulli-type trials with probability of success
`p`. In this particular case,
`n = 8`, and
`p = 0.53`, therefore, the model is
`{8 \choose x} (0.53) ^ {x} (0.47)^((8-x))`. So, you have:

`P (X = 0) = {8 \choose 0} (0.53) ^ {0} (0.47) ^ {8} = 0.0024`

`P (X = 1) = {8 \choose 1} (0.53) ^ {1} (0.47) ^ {7} = 0.0215`

`P (X = 2) = {8 \choose 2} (0.53)^2 (0.47)^6 = 0.0848`

`P (X = 3) = {8 \choose 3} (0.53) ^ {3} (0.47)^5 = 0.1912`

`P (X = 4) = {8 \choose 4} (0.53) ^ {4} (0.47)^4} = 0.2695`

`P (X = 5) = {8 \choose 5} (0.53) ^ {5} (0.47)^3 = 0.2431`

`P (X = 6) = {8 \choose 6} (0.53) ^ {6} (0.47)^2 = 0.1371`

`P (X = 7) = {8 \choose 7} (0.53) ^ {7} (0.47)^ {1} = 0.0442`

`P (X = 8) = {8 \choose 8} (0.53)^(8) (0.47)^(0) = 0.0062`

The unusual
`X` values ​​for this model are:
`X = 0, 1, 7, 8`