The distance that a Tesla model 3 can travel is normally distributed with a mean of 260 miles and a standard deviation of 25 miles. What is the probability that a randomly selected Tesla model 3 can travel more than 310 miles?

Respuesta :

Answer:

2.28% probability that a randomly selected Tesla model 3 can travel more than 310 miles

Step-by-step explanation:

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

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

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

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:

[tex]\mu = 260, \sigma = 25[/tex]

What is the probability that a randomly selected Tesla model 3 can travel more than 310 miles?

This is 1 subtracted by the pvalue of Z when X = 310. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{310 - 260}{25}[/tex]

[tex]Z = 2[/tex]

[tex]Z = 2[/tex] has a pvalue of 0.0228

2.28% probability that a randomly selected Tesla model 3 can travel more than 310 miles