Answer:
8.85% probability that fewer than 82 accounts in the sample were at least 1 month in arrears
Step-by-step explanation:
For each account, there are only two possible outcomes. Either they are at least 1 month in arrears, or they are not. The probability of an account being at least 1 month in arrears is independent from other accounts. So the binomial probability distribution is used to solve this question.
However, we are working with a large sample. So i am going to aproximate this binomial distribution to the normal.
Binomial probability distribution
Probability of exactly x sucesses 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:
![E(X) = np](https://tex.z-dn.net/?f=E%28X%29%20%3D%20np)
The standard deviation of the binomial distribution is:
![\sqrt{V(X)} = \sqrt{np(1-p)}](https://tex.z-dn.net/?f=%5Csqrt%7BV%28X%29%7D%20%3D%20%5Csqrt%7Bnp%281-p%29%7D)
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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 pvalue, 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
,
.
In this problem, we have that:
![p = 0.22, n = 425](https://tex.z-dn.net/?f=p%20%3D%200.22%2C%20n%20%3D%20425)
So
![\mu = E(X) = np = 425*0.22 = 93.5](https://tex.z-dn.net/?f=%5Cmu%20%3D%20E%28X%29%20%3D%20np%20%3D%20425%2A0.22%20%3D%2093.5)
![\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{425*0.22*0.78} = 8.54](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%7B425%2A0.22%2A0.78%7D%20%3D%208.54)
What is the probability that fewer than 82 accounts in the sample were at least 1 month in arrears
This probability is the pvalue of Z when X = 82. 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{82 - 93.5}{8.54}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B82%20-%2093.5%7D%7B8.54%7D)
![Z = -1.35](https://tex.z-dn.net/?f=Z%20%3D%20-1.35)
has a pvalue of 0.0885.
8.85% probability that fewer than 82 accounts in the sample were at least 1 month in arrears