Answer:
The outcomes of this binomial distribution that would be considered unusual is {0, 1, 2, 8}.
Explanation:
The outcomes provided are:
(A) 0, 1, 2, 6, 7, 8
(B) 0, 1, 2, 7, 8
(C) 0, 1, 7, 8
(D) 0, 1, 2, 8
Solution:
The random variable X can be defined as the number of employees who judge their co-workers by cleanliness.
The probability of X is:
P (X) = 0.65
The number of employees selected is:
n = 8
An unusual outcome, in probability theory, has a probability of occurrence less than or equal to 0.05.
Since outcomes 0 and 1 are contained in all the options, we will check for X = 2.
Compute the value of P (X = 2) as follows:
![P(X=2)={8\choose 2}(0.65)^(2)(0.35)^(8-2)](https://img.qammunity.org/2021/formulas/mathematics/high-school/51mdlpkava74rtggbmot1dats4r7efri0d.png)
![=28* 0.4225* 0.001838265625\\=0.02175\\\approx 0.022<0.05](https://img.qammunity.org/2021/formulas/mathematics/high-school/7uz0umqbmv76xfihehkf45kvxap7ud9k9g.png)
So X = 2 is unusual.
Similarly check for X = 6, 7 and 8.
P (X = 6) = 0.2587 > 0.05
X = 6 not unusual
P (X = 7) = 0.1373 > 0.05
X = 7 not unusual
P (X = 8) = 0.0319
X = 8 is unusual.
Thus, the outcomes of this binomial distribution that would be considered unusual is {0, 1, 2, 8}.