Question

The size of the left upper chamber of the heart is one measure of cardiovascular health. When the upper left chamber is enlarged, the risk of heart problems is increased. A paper described a study in which the left atrial size was measured for a large number of children age 5 to 15 years. Based on this data, the authors concluded that for healthy children, left atrial diameter was approximately normally distributed with a mean of 26.2 mm and a standard deviation of 4.3 mm. (a) Approximately what proportion of healthy children have left atrial diameters l

Answer 1

a)
`P(X<24)=0.304`

b)
`P(X>32)=0.0887`

c)
`P(25<X<30)=0.422`

d)
`X=26.2 + 0.842(4.3)=29.821`

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:

`X \sim N(26.2,4.3)`

Where
`\mu=26.2` and
`\sigma=4.3`

We are interested on this probability

`P(X<24)`

And the best way to solve this problem is using the normal standard distribution and the z score given by:

`z=(x-\mu)/(\sigma)`

If we apply this formula to our probability we got this:

`P(X<24)=P((X-\mu)/(\sigma)<(24-\mu)/(\sigma))=P(Z<(24-26.2)/(4.3))=P(Z<-0.512)`

And we can find this probability using excel or a table, on this way:

`P(Z<-0.512)=0.304`

3) Part b : Approximately what proportion of healthy children have left atrial diameters greater than 32 mm

`P(X>32)`

If we apply this formula to our probability we got this:

`P(X>32)=P((X-\mu)/(\sigma)>(32-\mu)/(\sigma))=P(Z>(32-26.2)/(4.3))=P(Z>1.349)`

We can use the complement rule and we can find this probability using excel or a table, on this way:

`P(Z>1.349)=1-P(Z<1.349)=0.0887`

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:

`P(25<X<30)=P((25-\mu)/(\sigma)<(X-\mu)/(\sigma)<(30-\mu)/(\sigma))=P((25-26.2)/(4.3)<Z<(30-26.2)/(4.3))=P(-0.279<z<0.884)`

And we can find this probability on this way:

`P(-0.279<z<0.884)=P(z<0.884)-P(z<-0.279)`

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.

`P(-0.279<z<0.884)=P(z<0.884)-P(z<-0.279)=0.812-0.390=0.422`

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

`z=(x-\mu)/(\sigma)`

If we solve for X w got:

`X=26.2 + 0.842(4.3)=29.821`