Answer:
The probability of getting a sample with 80% satisfied customers or less is 0.0125.
Step-by-step explanation:
We are given that the results of 1000 simulations, each simulating a sample of 80 customers, assuming there are 90 percent satisfied customers.
Let [tex]\hat p[/tex] = sample proportion of satisfied customers
The z-score probability distribution for the sample proportion is given by;
Z = [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ~ N(0,1)
where, p = population proportion of satisfied customers = 90%
n = sample of customers = 80
Now, the probability of getting a sample with 80% satisfied customers or less is given by = P( [tex]\hat p \leq[/tex] 80%)
P( [tex]\hat p \leq[/tex] 80%) = P( [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] [tex]\leq[/tex] [tex]\frac{0.80-0.90}{\sqrt{\frac{0.80(1-0.80)}{80} } }[/tex] ) = P(Z [tex]\leq[/tex] -2.24) = 1 - P(Z < 2.24)
= 1 - 0.9875 = 0.0125
The above probability is calculated by looking at the value of x = 2.24 in the z table which has an area of 0.9875.
