Respuesta :
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- 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 [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
In this problem:
- Mean of 62 mph, hence [tex]\mu = 62[/tex].
- Standard deviation of 5 mph, hence [tex]\sigma = 5[/tex].
- Sample of 50 cards, hence [tex]n = 50, s = \frac{5}{\sqrt{50}} = 0.7071[/tex]
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:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{66 - 62}{0.7071}[/tex]
[tex]Z = 5.66[/tex]
[tex]Z = 5.66[/tex] 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 https://brainly.com/question/24663213
