Ted likes to run long distances. He can run 20 \text{ km}20 km20, start text, space, k, m, end text in 959595 minutes. He wants
1 answer:
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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:
The standard deviation of the binomial distribution is:
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:
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:
So
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
has a pvalue of 0.0885.
8.85% probability that fewer than 82 accounts in the sample were at least 1 month in arrears