The unusual outcomes for the binomial distribution are given as follows:
0 or 1 employee making judgements.
What is the binomial probability distribution?
It is the probability of exactly x successes on n repeated trials, with p probability of a success on each trial.
The expected value of the binomial distribution is:
E(X) = np
The standard deviation of the binomial distribution is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]
In the binomial distribution, a measure is considered unusual if it is more than 2.5 standard deviations from the mean.
For this problem, the parameters are given as follows:
n = 8, p = 0.61.
Hence the mean and the standard deviation are given by:
- E(X) = np = 8 x 0.61 = 4.88.
- [tex]\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{8(0.61)(0.39)} = 1.38[/tex]
The bounds of usual outcomes are:
- 4.88 - 2.5 x 1.38 = 1.43.
- 4.88 + 2.5 x 1.38 = 8.33.
Hence outcomes of less than 1 employee making judgements would be unusual.
More can be learned about the binomial distribution at https://brainly.com/question/24863377
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