Question

Speeding on the I-5, Part I. The distribution of passenger vehicle speeds traveling on the Interstate 5 Freeway (I-5) in California is nearly normal with a mean of 72.6 miles/hour and a standard deviation of 4.78 miles/hour.47 (a) What percent of passenger vehicles travel slower than 80 miles/hour?(b) What percent of passenger vehicles travel between 60 and 80 miles/hour?(c) How fast to do the fastest 5% of passenger vehicles travel?(d) The speed limit on this stretch of the I-5 is 70 miles/hour. Approximate what percentage of the passenger vehicles travel above the speed limit on this stretch of the I-5.

Answer 1

a)
`P(X<80)=P((X-\mu)/(\sigma)<(80-\mu)/(\sigma))=P(Z<(80-72.6)/(4.78))=P(z<1.548)`

And we can find this probability using the normal standard distirbution or excel and we got:

`P(z<1.548)=0.939`

And that correspond to 93.9 %

b)
`P(60<X<80)=P((60-\mu)/(\sigma)<(X-\mu)/(\sigma)<(80-\mu)/(\sigma))=P((60-72.6)/(4.78)<Z<(80-72.6)/(4.78))=P(-2.636<z<1.548)`

And we can find this probability with this difference:

`P(-2.636<z<1.548)=P(z<1.548)-P(z<-2.636)`

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.

`P(-2.636<z<1.548)=P(z<1.548)-P(z<-2.636)=0.939-0.0042=0.935`

So we have approximately 93.5%

c)
`z=1.64<(a-72.6)/(4.78)`

And if we solve for a we got

`a=72.6 +1.64*4.78=80.439`

So the value of velocity that separates the bottom 95% of data from the top 5% is 80.439.

d)
`P(X>70)=P((X-\mu)/(\sigma)>(70-\mu)/(\sigma))=P(Z>(70-72.6)/(4.78))=P(z>-0.544)`

And we can find this probability using the complement rule, normal standard distirbution or excel and we got:

`P(z>-0.544)=1-P(z<-0.544) = 1-0.293=0.707`

And that correspond to 70.7 %

Explanation:

Previous concepts

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".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Part a

Let X the random variable that represent the vehicles speeds of a population, and for this case we know the distribution for X is given by:

`X \sim N(72.6,4.78)`

Where
`\mu=72.6` and
`\sigma=4.78`

We are interested on this probability

`P(X<80)`

And the best way to solve this problem is using the normal standard distribution and the z score given by:

`z=(x-\mu)/(\sigma)`

If we apply this formula to our probability we got this:

`P(X<80)=P((X-\mu)/(\sigma)<(80-\mu)/(\sigma))=P(Z<(80-72.6)/(4.78))=P(z<1.548)`

And we can find this probability using the normal standard distirbution or excel and we got:

`P(z<1.548)=0.939`

And that correspond to 93.9 %

Part b

We want this probability

`P(60<X<80)=P((60-\mu)/(\sigma)<(X-\mu)/(\sigma)<(80-\mu)/(\sigma))=P((60-72.6)/(4.78)<Z<(80-72.6)/(4.78))=P(-2.636<z<1.548)`

And we can find this probability with this difference:

`P(-2.636<z<1.548)=P(z<1.548)-P(z<-2.636)`

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.

`P(-2.636<z<1.548)=P(z<1.548)-P(z<-2.636)=0.939-0.0042=0.935`

So we have approximately 93.5%

Part c

For this part we want to find a value a, such that we satisfy this condition:

`P(X>a)=0.05` (a)

`P(X<a)=0.95` (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.95 of the area on the left and 0.05 of the area on the right it's z=1.64. On this case P(Z<1.64)=0.95 and P(z>1.64)=0.05

If we use condition (b) from previous we have this:

`P(X<a)=P((X-\mu)/(\sigma)<(a-\mu)/(\sigma))=0.95`

`P(z<(a-\mu)/(\sigma))=0.95`

But we know which value of z satisfy the previous equation so then we can do this:

`z=1.64<(a-72.6)/(4.78)`

And if we solve for a we got

`a=72.6 +1.64*4.78=80.439`

So the value of velocity that separates the bottom 95% of data from the top 5% is 80.439.

Part d

`P(X>70)`

And the best way to solve this problem is using the normal standard distribution and the z score given by:

`z=(x-\mu)/(\sigma)`

If we apply this formula to our probability we got this:

`P(X>70)=P((X-\mu)/(\sigma)>(70-\mu)/(\sigma))=P(Z>(70-72.6)/(4.78))=P(z>-0.544)`

And we can find this probability using the complement rule, normal standard distirbution or excel and we got:

`P(z>-0.544)=1-P(z<-0.544) = 1-0.293=0.707`

And that correspond to 70.7 %