Question

The number of accidents per week at a hazardous intersection varies with mean 2.2 and standard deviation 1.4. The distribution of this particular variable is very right skewed. (a) Suppose we let X be the mean number of accidents per week at the intersection during 9 randomly chosen weeks. What is the probability that X is less than 2?

Answer 1

Answer:

33.36% probability that X is less than 2.

Step-by-step explanation:

The distribution is not normal, however, using the central limit theorem, it is going to be approximately normal. So

Central Limit Theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$, a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

Normal Probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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 pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 2.2, \sigma = 1.4, n = 9, s = \frac{1.4}{\sqrt{9}} = 0.467$

(a) Suppose we let X be the mean number of accidents per week at the intersection during 9 randomly chosen weeks. What is the probability that X is less than 2?

This is the pvalue of Z when X = 2. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{2 - 2.2}{0.467}$

$Z = -0.43$

$Z = -0.43$ has a pvalue of 0.3336.

33.36% probability that X is less than 2.