Question

The number of accidents per week at a hazardous intersection is a random variable with mean 6.3 and standard deviation 5.85. The distribution of the number of accidents is very right skewed. (a) Suppose we let X be the sample average number of accidents per week at the intersection during 9 randomly chosen weeks. What is the probability that X is less than 5

Answer 1

The probability that X is less than 5 cannot be determined.

Explanation:

Normal Probability Distribution:

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

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

`Z = (X - \mu)/(\sigma)`

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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean
`\mu` and standard deviation
`\sigma`, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question:

The distribution is right-skewed, which means that the central limit theorem can only be applied for a sample size of at least 30. Since the sample size is 9 < 30, the CLT cannot be applied, and thus the probability that X is less than 5 cannot be determined.