## In a study investigation car speed in automobile accidents, 5000 accidents were examined and the vehicle speed at impact was recorded. Suppo

Question

In a study investigation car speed in automobile accidents, 5000 accidents were examined and the vehicle speed at impact was recorded. Suppose the average speed at impact in the study was 42 mph with a standard deviation of 15 mph. A histogram of the data revealed that the distribution of speeds was approximately normal (symmetric bell curve).a.Approximately what percent of the vehicle speeds were less than 42 mph

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2022-01-01T18:39:04+00:00
2022-01-01T18:39:04+00:00 2 Answers
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## Answers ( )

Answer:[tex] P(X<42)[/tex]

Since the sample size is large enough and we know that the distribution for X is approximately normal we can use the z score formula given by:

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

And for X =42 we got:

[tex] z = \frac{42-42}{15}= 0[/tex]

And we have this:

[tex] P(z<0) = 0.5[/tex]

So we expect abput 50% of the values below 42 mph.

Step-by-step explanation:Previous conceptsThe

central limit theoremstates that “if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large”.Normal distribution, is a “probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean”.Solution to the problemWe assume that the random variable X represent the speed at impact and we know from the sample size of n =5000 that:

[tex] E(X)= 42, \sigma =15[/tex]

And we want to find thie probability:

[tex] P(X<42)[/tex]

Since the sample size is large enough and we know that the distribution for X is approximately normal we can use the z score formula given by:

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

And for X =42 we got:

[tex] z = \frac{42-42}{15}= 0[/tex]

And we have this:

[tex] P(z<0) = 0.5[/tex]

So we expect abput 50% of the values below 42 mph.

Answer:Approximately 50% of the vehicle speeds were less than 42 mph

Step-by-step explanation:The normal distribution is symmetric:That is, 50% of the measures are below the mean and 50% of the measures are above the mean.

In this problem, we have that:Mean = 42 mph

So

Approximately 50% of the vehicle speeds were less than 42 mph