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Based on a recent study, the pH level of the arterial cord (one vessel in the umbilical cord) is normally distributed with me

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Answer #1

Let X : pH level of arterial cord is normally distributed with

=7.26 and =0.13

a) we asked P( 7.00 < X < 7.50)

= P( (7.00-7.26)/0.13 < X-/ <( 7.50-7.26)/0.13)

= P( -2 < Z < 1.85)

= P( Z < 1.85) - P(Z<-2)

= 0.9687 - 0.0228

P( 7.00 < X < 7.50) = 0.945

Percentage = 94.50%

b)

P( X > 7.34) = P( X -/ > (7.34-7.26)/0.13)

= P( Z > 0.62)

= 1- P( Z < 0.62)

= 1 - 0.7324

P( X>7.34) = 0.2676

Percentage = 26.76%

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