Question

Birth weights in the United States are normally distributed with a mean of 3420 g and a standard deviation of 495 g. The Newport General Hospital requires special treatment for babies that are less than 2450 g (unusually light) or more than 4390 g (unusually heavy). What is the percentage of babies who do not require special treatment because they have birth weights between 2450 g and 4390 g

Answer 1

Answer:

$P(2450<X<4390)=P(\frac{2450-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{4390-\mu}{\sigma})=P(\frac{2450-3420}{495}<Z<\frac{4390-3420}{495})=P(-1.96<z<1.96)$

And we can find this probability with this difference:

$P(-1.96<z<1.96)=P(z<1.96)-P(z<-1.96)$

If we use the normal standard table or excel we got:

$P(-1.96<z<1.96)=P(z<1.96)-P(z<-1.96)=0.975-0.025=0.95$

And that represent 95% of the data. so then the percentage below 2450 is 2.5% and above 4390 is 2.5 %

Step-by-step explanation:

Let X the random variable that represent the birth weights of a population, and for this case we know the distribution for X is given by:

$X \sim N(3420,495)$  

Where $\mu=3420$ and $\sigma=495$

We are interested on this probability

$P(2450<X<4390)$

And we can use the z score formula given byÑ

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(2450<X<4390)=P(\frac{2450-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{4390-\mu}{\sigma})=P(\frac{2450-3420}{495}<Z<\frac{4390-3420}{495})=P(-1.96<z<1.96)$

And we can find this probability with this difference:

$P(-1.96<z<1.96)=P(z<1.96)-P(z<-1.96)$

If we use the normal standard table or excel we got:

$P(-1.96<z<1.96)=P(z<1.96)-P(z<-1.96)=0.975-0.025=0.95$

And that represent 95% of the data. so then the percentage below 2450 is 2.5% and above 4390 is 2.5 %