Respuesta :
Answer:
a) [tex]P(X<24)=0.304[/tex]
b) [tex]P(X>32)=0.0887[/tex]
c) [tex]P(25<X<30)=0.422[/tex]
d) [tex]X=26.2 + 0.842(4.3)=29.821[/tex]
Step-by-step explanation:
1) 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".
2) Part a : Approximately what proportion of healthy children have left atrial diameters less than 24 mm
Let X the random variable that represent the size of the left upper chamber of the heart of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(26.2,4.3)[/tex]
Where [tex]\mu=26.2[/tex] and [tex]\sigma=4.3[/tex]
We are interested on this probability
[tex]P(X<24)[/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(X<24)=P(\frac{X-\mu}{\sigma}<\frac{24-\mu}{\sigma})=P(Z<\frac{24-26.2}{4.3})=P(Z<-0.512)[/tex]
And we can find this probability using excel or a table, on this way:
[tex]P(Z<-0.512)=0.304[/tex]
3) Part b : Approximately what proportion of healthy children have left atrial diameters greater than 32 mm
[tex]P(X>32)[/tex]
If we apply this formula to our probability we got this:
[tex]P(X>32)=P(\frac{X-\mu}{\sigma}>\frac{32-\mu}{\sigma})=P(Z>\frac{32-26.2}{4.3})=P(Z>1.349)[/tex]
We can use the complement rule and we can find this probability using excel or a table, on this way:
[tex]P(Z>1.349)=1-P(Z<1.349)=0.0887[/tex]
4) Part c : Approximately what proportion of healthy children have left atrial diameters between 25 and 30 mm
If we apply this formula to our probability we got this:
[tex]P(25<X<30)=P(\frac{25-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{30-\mu}{\sigma})=P(\frac{25-26.2}{4.3}<Z<\frac{30-26.2}{4.3})=P(-0.279<z<0.884)[/tex]
And we can find this probability on this way:
[tex]P(-0.279<z<0.884)=P(z<0.884)-P(z<-0.279)[/tex]
And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.
[tex]P(-0.279<z<0.884)=P(z<0.884)-P(z<-0.279)=0.812-0.390=0.422[/tex]
5) Part d: For healthy children, what is the value for which only about 20% have a larger left atrial diameter?
For this case we want a value that accumulates 0.2 of the area on th right tail of the distribution. And for this we can use the Z score in order fo find the X value.
First we find a z score that accumulates 0.2 of the area on the right tail and 0.8 of the area on the left, and this value is z=0.842
[tex]z=\frac{x-\mu}{\sigma}[/tex]
If we solve for X w got:
[tex]X=26.2 + 0.842(4.3)=29.821[/tex]
