Respuesta :
Answer:
a) [tex]P(X<50)=P(\frac{X-\mu}{\sigma}<\frac{50-\mu}{\sigma})=P(Z<\frac{50-46.8}{1.75})=P(z<1.829)[/tex]
And we can find this probability using the normal standard table or excel and we got:
[tex]P(z<1.829)=0.966[/tex]
b) [tex]P(X>48)=P(\frac{X-\mu}{\sigma}>\frac{48-\mu}{\sigma})=P(Z>\frac{48-46.8}{1.75})=P(z>0.686)[/tex]
And we can find this probability using the complement rule, the normal standard table or excel and we got:
[tex]P(z>0.686)=1- P(Z<0.686)= 1-0.754=0.246[/tex]
c) [tex]P(46.8-1.5*1.75<X<46.8+1.5*1.75)=P(\frac{44.175-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{49.425-\mu}{\sigma})=P(\frac{44.175-46.8}{1.75}<Z<\frac{49.425-46.8}{1.75})=P(-1.5<z<1.5)[/tex]
And we can find this probability with this difference:
[tex]P(-1.5<z<1.5)=P(z<1.5)-P(z<-1.5)[/tex]
And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.
[tex]P(-1.5<z<1.5)=P(z<1.5)-P(z<-1.5)=0.933-0.0668=0.866[/tex]
Step-by-step 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 velocities of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(46.8,1.75)[/tex]
Where [tex]\mu=46.8[/tex] and [tex]\sigma=1.75[/tex]
We are interested on this probability
[tex]P(X<50)[/tex]
And the best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
If we apply this formula to our probability we got this:
[tex]P(X<50)=P(\frac{X-\mu}{\sigma}<\frac{50-\mu}{\sigma})=P(Z<\frac{50-46.8}{1.75})=P(z<1.829)[/tex]
And we can find this probability using the normal standard table or excel and we got:
[tex]P(z<1.829)=0.966[/tex]
Part b
[tex]P(X>48)=P(\frac{X-\mu}{\sigma}>\frac{48-\mu}{\sigma})=P(Z>\frac{48-46.8}{1.75})=P(z>0.686)[/tex]
And we can find this probability using the complement rule, the normal standard table or excel and we got:
[tex]P(z>0.686)=1- P(Z<0.686)= 1-0.754=0.246[/tex]
Part c
[tex]P(46.8-1.5*1.75<X<46.8+1.5*1.75)=P(\frac{44.175-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{49.425-\mu}{\sigma})=P(\frac{44.175-46.8}{1.75}<Z<\frac{49.425-46.8}{1.75})=P(-1.5<z<1.5)[/tex]
And we can find this probability with this difference:
[tex]P(-1.5<z<1.5)=P(z<1.5)-P(z<-1.5)[/tex]
And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.
[tex]P(-1.5<z<1.5)=P(z<1.5)-P(z<-1.5)=0.933-0.0668=0.866[/tex]
