Question

Let X be the number of accidents per week at the Hillsborough Street roundabout by the NCSU Bell Tower. Assume X varies with mean 2.2 and standard deviation 1.4. Note that X has only small whole-number values so the distribution of X cannot be reasonably approximated with a normal model.

Question 1. Let x be the mean number of accidents per week at the roundabout during a year (52 weeks). What is the probability that x is less than 2? (Use 4 decimal places in your answer).

Question 2. What is the probability that there are fewer than 100 accidents at the roundabout in a year? (Use 4 decimal places in your answer).

Answer 1

1) 0.1587 = 15.15% probability that x is less than 2

2) 7.64% probability that there are fewer than 100 accidents at the roundabout in a year

Explanation:

To solve this question, we need to understand the normal probability distribution and the Central limit theorem.

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 = (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.

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 problem, we have that:

`\mu = 2.2, \sigma = 1.4`

1)

Here we have
`\mu = 2.2, s = (1.4)/(√(52)) = 0.194`

What is the probability that x is less than 2?

This is the pvalue of Z when X = 2.

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

By the Central Limit Theorem

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

`Z = (2 - 2.2)/(0.194)`

`Z = -1.03`

`Z = -1.03` has a pvalue of 0.1515

0.1587 = 15.15% probability that x is less than 2

2)

100/52 = 1.9231

This is the pvalue of Z when X = 1.9231

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

`Z = (1.9231 - 2.2)/(0.194)`

`Z = -1.43`

`Z = -1.43` has a pvalue of 0.0764

7.64% probability that there are fewer than 100 accidents at the roundabout in a year