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 26.7 mm and a standard deviation of 4.2 mm.

(a)

Approximately what proportion of healthy children have left atrial diameters less than 24 mm? (Round your answer to four decimal places.)

(b)

Approximately what proportion of healthy children have left atrial diameters greater than 32 mm? (Round your answer to four decimal places.)

(c)

Approximately what proportion of healthy children have left atrial diameters between 25 and 30 mm? (Round your answer to four decimal places.)

(d)

For healthy children, what is the value for which only about 20% have a larger left atrial diameter? (Round your answer to two decimal places.)

  

You may need to use the appropriate table in Appendix A to answer this question.

Answer 1

Mean = = 26.7

Standard deviation = = 4.2

a) We have to find the proportion of healthy children have left atrial diameters less than 24 mm.

That is we have to find P(X < 24)

For finding this probability we have to find z score.

That is we have to find P(Z < - 0.64)

P(Z < - 0.64) = 0.2611 ( Using z table)

So the proportion of healthy children have left atrial diameters less than 24 mm = 0.2611

b) We have to find the proportion of healthy children have left atrial diameters greater than 32 mm.

That is we have to find P(X > 32)

For finding this probability we have to find z score.

That is we have to find P(Z > 1.26)

P(Z > 1.26) = 1 - P(z < 1.26) = 1 - 0.8962 = 0.1038 ( Using z table)

So the proportion of healthy children have left atrial diameters greater than 32 mm is 0.1038

c) We have to find the proportion of healthy children have left atrial diameters between 25 and 30 mm.

That is we have to find P(25 < X < 30)

For finding this probability we have to find z score.

That is we have to find P(- 0.40 < Z > 0.79)

P( - 0.40 < Z > 0.79) = P(Z < 0.79) - P(Z < - 0.40) = 0.7852 - 0.3446 = 0.4406 ( Using z table)

So the proportion of healthy children have left atrial diameters between 25 and 30 mm is 0.4406

d) Now we have to the find what is the value for which only about 20% have a larger left atrial diameter.

That means we have to find the value of x.

First, we have to find z value for 20% has a larger left atrial.

So z = 0.84 ( Using z table)

So the value for which only about 20% have a larger left atrial diameter is 30.23