Question

Based on a recent​ study, the pH level of the arterial cord​ (one vessel in the umbilical​ cord) is normally distributed with mean 7.36 and standard deviation of 0.11. Find the percentage of preterm infants who have the following arterial cord pH levels.

a. pH levels between 7.00 and 7.50.

b. pH levels over 7.44.

Answer 1

Solution:

Given:  the pH level of the arterial cord​(one vessel in the umbilical​ cord) is normally distributed with mean 7.36 and standard deviation of 0.11.

Thus mean = = 7.36 and Standard deviation = = 0.11

Part a) Find the percentage of preterm infants who have  pH levels between 7.00 and 7.50

That is find:

P( 7.00 < X < 7.50) =..........?

Find z score for x = 7.00 and for x = 7.50

z score formula:

Thus we get:

P( 7.00 < X < 7.50) = P( -3.27 < Z < 1.27 )

P( 7.00 < X < 7.50) = P( Z < 1.27 ) - P( Z < -3.27 )

Look in z table for z = 1.2 and 0.07 as well as for z = -3.2 and 0.07  and find area.

P( Z < 1.27 ) = 0.8980

and

P( Z < -3.27)= 0.0005

P( 7.00 < X < 7.50) = P( Z < 1.27 ) - P( Z < -3.27 )

P( 7.00 < X < 7.50) = 0.8980 - 0.0005

P( 7.00 < X < 7.50) = 0.8975

P( 7.00 < X < 7.50) = 89.75%

Part b) the percentage of preterm infants who have  pH levels  over 7.44

P( X > 7.44) = ...........?

Find z score for x = 7.44

P( X > 7.44) = P( Z > 0.73 )

P( X > 7.44) = 1 - P( Z < 0.73 )

Look in z table for z = 0.7 and 0.03 and find area.

P( Z < 0.73 ) = 0.7673

Thus

P( X > 7.44) = 1 - P( Z < 0.73 )

P( X > 7.44) = 1 - 0.7673

P( X > 7.44) = 0.2327

P( X > 7.44) = 23.27%