Respuesta :
Answer:
[tex]P(221<X<237)=P(\frac{221-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{237-\mu}{\sigma})=P(\frac{221-190}{21}<Z<\frac{237-190}{21})=P(1.476<z<2.238)[/tex]
And we can find this probability with this difference:
[tex]P(1.476<z<2.238)=P(z<2.238)-P(z<1.476)[/tex]
And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.
[tex]P(1.476<z<2.238)=P(z<2.238)-P(z<1.476)=0.987-0.924=0.063[/tex]
Step-by-step explanation:
For this case we assume the following complete question: "Let x denote the time taken to run a road race. Suppose x is approximately normally distributed with a mean of 190 minutes and a standard deviation of 21 minutes. If one runner is selected at random, what is the probability that this runner will complete this road race in 221 to 237 minutes? "
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 time taken to run road race of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(190,21)[/tex]
Where [tex]\mu=190[/tex] and [tex]\sigma=21[/tex]
We are interested on this probability
[tex]P(221<X<237)[/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(221<X<237)=P(\frac{221-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{237-\mu}{\sigma})=P(\frac{221-190}{21}<Z<\frac{237-190}{21})=P(1.476<z<2.238)[/tex]
And we can find this probability with this difference:
[tex]P(1.476<z<2.238)=P(z<2.238)-P(z<1.476)[/tex]
And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.
[tex]P(1.476<z<2.238)=P(z<2.238)-P(z<1.476)=0.987-0.924=0.063[/tex]
