Answer:
[tex]P(34<X<63)=P(\frac{34-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{63-\mu}{\sigma})=P(\frac{34-50}{7}<Z<\frac{63-50}{7})=P(-2.286<z<1.857)[/tex]
And we can find this probability with this difference and using the normal standard table or excel:
[tex]P(-2.286<z<1.857)=P(z<1.857)-P(z<-2.286)=0.968-0.0111= 0.9569[/tex]
And the result is illustrated in the figure attached.
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 variable of interest of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(50,7)[/tex]
Where [tex]\mu=50[/tex] and [tex]\sigma=7[/tex]
We are interested on this probability
[tex]P(34<X<63)[/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(34<X<63)=P(\frac{34-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{63-\mu}{\sigma})=P(\frac{34-50}{7}<Z<\frac{63-50}{7})=P(-2.286<z<1.857)[/tex]
And we can find this probability with this difference and using the normal standard table or excel:
[tex]P(-2.286<z<1.857)=P(z<1.857)-P(z<-2.286)=0.968-0.0111= 0.9569[/tex]
And the result is illustrated in the figure attached.
