Answer:
0.7734 = 77.34% probability that at least 22 will contain an error. Probability above 50%, which means that this is likely to occur.
Step-by-step explanation:
Binomial probability distribution
Probability of exactly x successes on n repeated trials, with p probability.
Can be approximated to a normal distribution, using the expected value and the standard deviation.
The expected value of the binomial distribution is:
The standard deviation of the binomial distribution is:
Normal probability distribution
Problems of normally distributed distributions can be solved using the z-score formula.
In a set with mean
and standard deviation
, the z-score of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
When we are approximating a binomial distribution to a normal one, we have that
,
.
About 12.5% of restaurant bills are incorrect.
This means that ![p = 0.125](https://tex.z-dn.net/?f=p%20%3D%200.125)
200 bills are selected at random
This means that ![n = 200](https://tex.z-dn.net/?f=n%20%3D%20200)
Mean and standard deviation:
![\mu = E(X) = np = 200*0.125 = 25](https://tex.z-dn.net/?f=%5Cmu%20%3D%20E%28X%29%20%3D%20np%20%3D%20200%2A0.125%20%3D%2025)
![\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{200*0.125*0.875} = 4.677](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%5Csqrt%7BV%28X%29%7D%20%3D%20%5Csqrt%7Bnp%281-p%29%7D%20%3D%20%5Csqrt%7B200%2A0.125%2A0.875%7D%20%3D%204.677)
Find the probability that at least 22 will contain an error.
Using continuity correction, this is
, which is 1 subtracted by the p-value of Z when X = 21.5. So
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{21.5 - 25}{4.677}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B21.5%20-%2025%7D%7B4.677%7D)
![Z = -0.75](https://tex.z-dn.net/?f=Z%20%3D%20-0.75)
has a p-value of 0.2266.
1 - 0.2266 = 0.7734
0.7734 = 77.34% probability that at least 22 will contain an error. Probability above 50%, which means that this is likely to occur.