Sixty-four percent of those that use drive through services believe that the employees are at least somewhat rude. If you asked 87 people who use drive through services if they believe that the employees are at least somewhat rude, what is the probability that at most 60 say yes?

a. 0.860
b. 0.802
c. 0.057
d. 0.198

Answer:

The answer to this question would normally be, 85.96%, though that isnt an answer choice so the closest thing to it would most likely be answer choice A.

Using the binomial approximation to the normal, it is found that:

0.86 probability that at most 60 say yes, given by option a.

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Binomial probability distribution

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

The expected value is:

[tex]E(X) = np[/tex]

The standard deviation is:

[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]

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Normal probability distribution

Problems of normally distributed distributions can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

When we are approximating a binomial distribution to a normal one, we have that [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex], if [tex]np \geq 10[/tex] and [tex]n(1-p) \geq 10[/tex].

----------------------------

The mean and standard deviation are given by:

[tex]\mu = E(X) = np = 87(0.64) = 55.68[/tex]

[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{87(0.64)(0.36)} = 4.4771[/tex]

Using continuity correction, the probability that at most 60 use is [tex]P(X \leq 60 + 0.5) = P(X \leq 60.5)[/tex], which is the p-value of Z when X = 60.5, thus:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{60.5 - 55.68}{4.4771}[/tex]

[tex]Z = 1.08[/tex]

[tex]Z = 1.08[/tex] has a p-value of 0.86.

Thus, 0.86 probability that at most 60 say yes, given by option a.

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