Question

Based on a recent​ study, the pH level of the arterial cord​ (one vessel in the umbilical​ cord) is normally distributed with mean 7.31 and standard deviation of 0.16. 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 7.39.

Answer 1

For normal distribution, P(X < A) = P(Z < (A - mean)/standard deviation)

Mean = 7.31

Standard deviation = 0.16

a) P(pH levels between 7.00 and 7.50) = P(7.00 < X < 7.50)

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

= P(Z < (7.50 - 7.31)/0.16) - P(Z < (7.00 - 7.31)/0.16)

= P(Z < 1.19) - P(Z < -1.94)

= 0.8830 - 0.0262

= 0.8568

b) P(pH levels over 7.39) = P(X > 7.39)

= 1 - P(X < 7.39)

= 1 - P(Z < (7.39 - 7.31)/0.16)

= 1 - P(Z < 0.5)

= 1 - 0.6915

= 0.3085