Respuesta :
Answer:
[tex]P(57<X<59)=P(\frac{57-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{59-\mu}{\sigma})=P(\frac{57-61}{6}<Z<\frac{59-61}{6})=P(-0.67<Z<-0.33)[/tex]
[tex]P(-0.67<z<-0.33)=P(z<-0.33)-P(z<-0.67)[/tex]
[tex]P(-0.67<z<-0.33)=P(z<-0.33)-P(z<-0.67)=0.371-0.251=0.119[/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".
Solution to the problem
Let X the random variable that represent the lifetime for a TV of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(61,6)[/tex]
Where [tex]\mu=61[/tex] and [tex]\sigma=6[/tex]
We are interested on this probability
[tex]P(57<X<59)[/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(57<X<59)=P(\frac{57-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{59-\mu}{\sigma})=P(\frac{57-61}{6}<Z<\frac{59-61}{6})=P(-0.67<Z<-0.33)[/tex]
And we can find this probability like this:
[tex]P(-0.67<z<-0.33)=P(z<-0.33)-P(z<-0.67)[/tex]
And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.
[tex]P(-0.67<z<-0.33)=P(z<-0.33)-P(z<-0.67)=0.371-0.251=0.119[/tex]
