Respuesta :
Answer:
[tex]P(3<X<6)=P(\frac{3-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{6-\mu}{\sigma})=P(\frac{3-4.5}{2.3}<Z<\frac{6-4.5}{2.3})=P(-0.65<z<0.65)[/tex]
And we can find this probability with this difference:
[tex]P(-0.65<z<0.65)=P(z<0.65)-P(z<-0.65)[/tex]
And in order to find these probabilities we use tables for the normal standard distribution, excel or a calculator and we got.
[tex]P(-0.65<z<0.65)=P(z<0.65)-P(z<-0.65)=0.742-0.258=0.484[/tex]
Step-by-step explanation:
Asuming this complete question: Suppose the number of hours to complete this exam for my students is denoted by the random variable X and is normally distributed with the mean of µx hours and the standard deviation of σX = 2.3 hours. Find the probability that a randomly selected student has finished the exam spending 3 to 6 hours, P(3 < X < 6) =? (Use the attached table.)
Let =4.5
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 number of hours to complete an examn of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(4.5,2.3)[/tex]
Where [tex]\mu=4.5[/tex] and [tex]\sigma=2.3[/tex]
We are interested on this probability
[tex]P(3<X<6)[/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(3<X<6)=P(\frac{3-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{6-\mu}{\sigma})=P(\frac{3-4.5}{2.3}<Z<\frac{6-4.5}{2.3})=P(-0.65<z<0.65)[/tex]
And we can find this probability with this difference:
[tex]P(-0.65<z<0.65)=P(z<0.65)-P(z<-0.65)[/tex]
And in order to find these probabilities we use tables for the normal standard distribution, excel or a calculator and we got.
[tex]P(-0.65<z<0.65)=P(z<0.65)-P(z<-0.65)=0.742-0.258=0.484[/tex]
