Answer:
a
[tex]P(X < 2500) = 0.02668[/tex]
b
[tex]P(X < 1500) = 0.00001[/tex]
Step-by-step explanation:
From the question we are told that
The population mean is [tex]\mu = 3350[/tex]
The standard deviation is [tex]\sigma = 440[/tex]
We also told in the question that the birth weight is approximately Normally distributed
i.e [tex]X \ \~ \ N(\mu , \sigma )[/tex]
Given that Low-birth-weight babies weighing less than 2500 grams,then the proportion of babies born full term are low-birth-weight babies is mathematically represented as
[tex]P(X < 2500) = P(\frac{ X - \mu }{\sigma } < \frac{2500 - \mu}{\sigma } )[/tex]
Generally
[tex]\frac{X - \mu}{ \sigma } = Z (The \ standardized \ value \ of \ X )[/tex]
So
[tex]P(X < 2500) = P(Z < \frac{2500 - \mu}{\sigma } )[/tex]
substituting values
[tex]P(X < 2500) = P(Z < \frac{2500 - 3350}{440 } )[/tex]
[tex]P(X < 2500) = P(Z <-1.932 )[/tex]
Now from the standardized normal distribution table(These value can also be obtained from Calculator dot com) the value of
[tex]P(Z <-1.932 ) = 0.02668[/tex]
=> [tex]P(X < 2500) = 0.02668[/tex]
Given that very-low-birth-weight babies (weighing less than 1500 grams,then the proportion of babies born full term are very-low-birth-weight babies is mathematically represented as
[tex]P(X < 1500) = P(\frac{ X - \mu }{\sigma } < \frac{1500 - \mu}{\sigma } )[/tex]
[tex]P(X < 1500) = P(Z < \frac{1500 - \mu}{\sigma } )[/tex]
substituting values
[tex]P(X < 1500) = P(Z < \frac{1500 - 3350}{440 } )[/tex]
[tex]P(X < 1500) = P(Z <-4.205 )[/tex]
Now from the standardized normal distribution table(These value can also be obtained from calculator dot com) the value of
[tex]P(Z <-1.932 ) = 0.00001[/tex]
[tex]P(X < 1500) = 0.00001[/tex]
