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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 4.7 mm.

(a) Approximately what proportion of healthy children have left atrial diameters less than 24 mm?
(b) Approximately what proportion of healthy children have left atrial diameters greater than 32 mm?
(c) Approximately what proportion of healthy children have left atrial diameters between 25 and 30 mm?
(d) For healthy children, what is the value for which only about 20% have a larger left atrial diameter?

Respuesta :

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Answer:

a) 0.283  or 28.3%

b) 0.130 or 13%

c) 0.4 or 40%

d) 30.6 mm

Step-by-step explanation:

z-score of a single left atrial diameter value of healthy children can be calculated as:

z=[tex]\frac{X-M}{s}[/tex] where

  • X is the left atrial diameter value we are looking for its z-score
  • M is the mean left atrial diameter of healthy children (26.7 mm)
  • s is the standard deviation (4.7 mm)

Then

a) proportion of healthy children who have left atrial diameters less than 24 mm

=P(z<z*) where z* is the z-score of 24 mm

z*=[tex]\frac{24-26.7}{4.7}[/tex] ≈ −0.574

And P(z<−0.574)=0.283

b) proportion of healthy children who have left atrial diameters greater than 32 mm

= P(z>z*) = 1-P(z<z*) where z* is the z-score of 32 mm

z*=[tex]\frac{32-26.7}{4.7}[/tex] ≈ 1.128

1-P(z<1.128)=0.8703=0.130

c) proportion of healthy children have left atrial diameters between 25 and 30 mm

=P(z(25)<z<z(30)) where z(25), z(30) are the z-scores of 25 and 30 mm

z(30)=[tex]\frac{30-26.7}{4.7}[/tex] ≈ 0.702

z(25)=[tex]\frac{25-26.7}{4.7}[/tex] ≈ −0.362

P(z<0.702)=0.7587

P(z<−0.362)=0.3587

Then P(z(25)<z<z(30)) =0.7587 - 0.3587 =0.4

d) to find the value for which only about 20% have a larger left atrial diameter, we assume

P(z>z*)=0.2 or 20% where z* is the z-score of the value we are looking for.

Then P(z<z*)=0.8 and z*=0.84. That is

0.84=[tex]\frac{X-26.7}{4.7}[/tex]  solving this equation for X we get X=30.648

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