Question

(CO 3) Sixty-five 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? 0, 1,2, 7, 8 0, 1, 2, 6, 7, 8 0, 1, 2, 8 0, 1, 7, 8

Answer 1

Answer:

The 0 and 1 outcomes would be considered unusual.

Step-by-step explanation:

We can define as usual outcomes, the ones that fall between $\bar{x}\pm 3s$.

According to a binomial distribution, s value is

$s=\sqrt{n*p*(1-p)}\\s=\sqrt{8*0.65*(1-0.65)}\\s=\sqrt{1.82}=1.35$

The expected value out of a 8 people sample is

$\bar{x}=p*n=0.65*8=5.2$

Then, our interval of usual values lies between

$\bar{x}- 3s\leq x \leq \bar{x} +3s\\\\5.20-3*1.35\leq x \leq 5.20+3*1.35\\1.15\leq x \leq 9.25$

We can conclude that 0 and 1 (the results that lie outside of the interval) are unusual.