Answer :
Answer:
[tex]P(25<X<34)=P(\frac{25-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{34-\mu}{\sigma})=P(\frac{25-25}{6}<Z<\frac{34-25}{6})=P(0<z<1.5)[/tex]
And we can find this probability with this difference:
[tex]P(0<z<1.5)=P(z<1.5)-P(z<0)[/tex]
And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.
[tex]P(0<z<1.5)=P(z<1.5)-P(z<0)=0.933-0.5=0.433[/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 hous spent studying the week before final exams of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(25,6)[/tex]
Where [tex]\mu=25[/tex] and [tex]\sigma=6[/tex]
We are interested on this probability
[tex]P(25<X<34)[/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(25<X<34)=P(\frac{25-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{34-\mu}{\sigma})=P(\frac{25-25}{6}<Z<\frac{34-25}{6})=P(0<z<1.5)[/tex]
And we can find this probability with this difference:
[tex]P(0<z<1.5)=P(z<1.5)-P(z<0)[/tex]
And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.
[tex]P(0<z<1.5)=P(z<1.5)-P(z<0)=0.933-0.5=0.433[/tex]