Question

*Based on a recent study, the pH level of the arterial cord (one vessel in the umbilical cord) is normally distributed with mean 7.21 and standard deviation of 0.15. Find the percentageof preterm infants who have the following arterial cord pH levels.a. pH levels between 7.00 and 7.50.b. pH levels over 7.29.a. The percentage of arterial cord pH levels that are between 7.00 and 7.50 is 0.89%(Round to two decimal places as needed.)b. The percentage of arterial cord pH levels that are over 7.29 is %(Round to two decimal places as needed.)

Answer 1

Given:

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

The mean = μ = 7.21

The standard deviation = σ = 0.15

For the required, we will use the following formula to use the z-score:

`z=(x-\mu)/(\sigma)`

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Find the percentage of preterm infants who have the following arterial cord pH levels.

a. pH levels between 7.00 and 7.50.

So, we will find the value of z-score when x = 7 and when x = 7.5

`\begin{gathered} x=7\to z=(7-7.21)/(0.15)=-1.4 \\ \\ x=7.5\to z=(7.5-7.21)/(0.15)=1.933 \end{gathered}`

So, we will find the probability of P( -1.4 < z < 1.933 ) from the z-tables

`P(-1.4The answer as a percentage = <strong>89.26%</strong><p>==================================================================</p><p> Find the percentage of preterm infants who have the following arterial cord pH levels. b. pH levels over 7.29.</p><p></p><p>So, we will find the value of the z-score when x = 7.29</p>[tex]x=7.29\to z=(7.29-7.21)/(0.15)=0.533`

So, we will find the probability of P ( z > 0.533 )from the z-tables

`P(z>0.533)=0.2969`

So, the answer as a percentage = 29.69%

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