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= where
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*= ≈ −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*= ≈ 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)= ≈ 0.702
z(25)= ≈ −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= solving this equation for X we get X=30.648
To isolate m, we'll multiply both sides by 8.
We go from m/8 < 4 to m < 32
This describes any number smaller than 32. We aren't including 32 itself, so we have an <u>open circle</u> at 32 on the number line. The <u>shading is to the left</u> since everything to the left of 32 is smaller than 32.
