Question

Mopeds (small motorcycles) are very popular in Europe because of their mobility, ease of operation, and low cost. An article2 described a rolling bench test for determining maximum vehicle speed. A normal distribution with mean value 46.8 km/h and standard deviation 1.75 km/h is postulated. Consider randomly selecting a single such moped.

Answer 1

a)
`P(X<50)=P((X-\mu)/(\sigma)<(50-\mu)/(\sigma))=P(Z<(50-46.8)/(1.75))=P(z<1.829)`

And we can find this probability using the normal standard table:

`P(z<1.829)=0.966`

b)
`P(X>48)=P((X-\mu)/(\sigma)>(48-\mu)/(\sigma))=P(Z>(48-46.8)/(1.75))=P(Z>0.686)=1-P(z<0.686)`

And we can find this probability using the normal standard table and the complement ruel:

`P(Z>0.686)=1-P(z<0.686)=1-0.754=0.246`

c)
`P(46.8-1.5*1.75<X<46.8+1.5*1.75)=P((44.175-\mu)/(\sigma)<(X-\mu)/(\sigma)<(49.425-\mu)/(\sigma))=P((44.175-46.8)/(2.6)<Z<(49.425-46.8)/(2.6))=P(-1.009<z<1.009)`

And we can find this probability with this difference:

`P(-1.009<z<1.009)=P(z<1.009)-P(z<-1.009)`

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

`P(-1.009<z<1.009)=P(z<1.009)-P(z<-1.009)=0.844-0.156=0.687`

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".

Assuming the following questions:

Part a: What is the probability that the maximum speed is at most 50 km/h?

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

`X \sim N(46.8,1.75)`

Where
`\mu=46.8` and
`\sigma=1.75`

We are interested on this probability

`P(X<50)`

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<50)=P((X-\mu)/(\sigma)<(50-\mu)/(\sigma))=P(Z<(50-46.8)/(1.75))=P(z<1.829)`

And we can find this probability using the normal standard table:

`P(z<1.829)=0.966`

Part b: What is the probability that maximum speed is at least 48 km/h?

`P(X>48)=P((X-\mu)/(\sigma)>(48-\mu)/(\sigma))=P(Z>(48-46.8)/(1.75))=P(Z>0.686)=1-P(z<0.686)`

And we can find this probability using the normal standard table and the complement ruel:

`P(Z>0.686)=1-P(z<0.686)=1-0.754=0.246`

Part c:What is the probability that maximum speed differs from the mean value by at most 1.5 standard deviations?

`P(46.8-1.5*1.75<X<46.8+1.5*1.75)=P((44.175-\mu)/(\sigma)<(X-\mu)/(\sigma)<(49.425-\mu)/(\sigma))=P((44.175-46.8)/(2.6)<Z<(49.425-46.8)/(2.6))=P(-1.009<z<1.009)`

And we can find this probability with this difference:

`P(-1.009<z<1.009)=P(z<1.009)-P(z<-1.009)`

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

`P(-1.009<z<1.009)=P(z<1.009)-P(z<-1.009)=0.844-0.156=0.687`