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

Answer:

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

Step-by-step 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: $X = 0, 1, 7, 8$

Answer 2

Answer:

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

Step-by-step 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:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

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

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

$\sqrt{V(X)} = \sqrt{np(1-p)} = 1.41$

$4.24 - 2.5*1.41 = 0.71$

$4.24 + 2.5*1.41 = 7.77$

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