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 26.8 mm and a standard deviation of 4.8 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.)

mm

Answer 1

Given:

Let x = Healthy children, left atrial diameter, which follows Normal distribution with mean µ = 26.8   and σ = 4.8

Part a)

x= Healthy children, left atrial diameter.

We have to find proportion of Healthy children, left atrial diameter less than 24.

We can write it as P(x < 24)

As x is a random variable with mean µ = 26.8   and σ = 4.8

First we will convert normal distribution to standard normal distribution using formula

Z =  

P(x < 24) = P(Z < )

                    =P(   Z <    )

=P(Z < -0.58)                                  Using normal probability table for left side we write

= 0.2810

You can also find this using excel command “=NORMSDIST(-0.58) “

Proportion of healthy children have left atrial diameter less than 24 is 0.2810

Part b)

We have to find Healthy children, left atrial diameter is greater than 32.

We can write it as P(x > 32)

P(x > 32) = P(Z > )

                    =P(   Z > )

=P(Z > 1.08)                                  Using normal probability table for left side we write

= 1 – P( Z < 1.08)

You can also find this using excel command “=NORMSDIST(1.08) “

=1- 0.8599

P(x > 32) = 0.1401

Proportion of healthy children have left atrial diameter greater than 32 is 0.1401

Part c)

We have to find Proportion of healthy children have left atrial diameter between 25 and 30

We can write it as P( 25 < x < 30)

P(25 < x < 30) = P( < Z < )

                    =P(      < Z < )

=P(-0.38 < Z < 0.66)   

= P (Z < 0.66) – P(Z < -0.38)         using normal probability table for left side

=0.7454 – 0.3520

You can also find this using excel commands “=NORMSDIST(0.66) “ and “=NORMSDIST(-0.38) “

P(25 < x < 30)= 0.3934

Proportion of healthy children have left atrial diameter between 25 and 30 is 0.3934

Part d)

We have to find value of x for which only 20% have a larger left atrial diameter.

First, we have to find value of z such that 20% of area under the standard normal curve lies to the right of z.

20 % area to the right of z = 1 – 80% area to the left of z.

Using normal probability table we have to find value of z for probability 0.80, so z = 0.84

You can also find this using excel commands “=NORMSINV(0.8) “

Using z =     , solve it for x

0.84 =

0.84*4.8 = x – 26.8

x = 30.832

For healthy children 30.83 is the diameter value for which about 20% have a larger atrial diameter.