Question

On a particular stretch of highway, the State Police know that the average speed is 62 mph with a standard deviation of 5 mph. On a busy holiday weekend, the police are concerned that people travel too fast. So they randomly monitor speeds of a sample of 50 cars and record an average speed of 66 mph. Use central limit theorem to calculate

Answer 1

Using the normal distribution and the central limit theorem, it is found that there is a 0% probability of a sample of 50 cars recording an average speed of 66 mph or higher.

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

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

• It 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, which is the percentile of X.
• By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

In this problem:

• Mean of 62 mph, hence $\mu = 62$.
• Standard deviation of 5 mph, hence $\sigma = 5$.
• Sample of 50 cards, hence $n = 50, s = \frac{5}{\sqrt{50}} = 0.7071$

The probability of a sample of 50 cars recording an average speed of 66 mph or higher is 1 subtracted by the p-value of Z when X = 66, hence:

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

By the Central Limit Theorem

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

$Z = \frac{66 - 62}{0.7071}$

$Z = 5.66$

$Z = 5.66$ has a p-value of 1.

1 - 1 = 0.

There is a 0% probability of a sample of 50 cars recording an average speed of 66 mph or higher.

A similar problem is given at brainly.com/question/24663213