Question

The weights of newborn baby boys born at a local hospital are believed to have a normal distribution with a mean weight of 3946 grams and a variance of 368,449. If a newborn baby boy born at the local hospital is randomly selected, find the probability that the weight will be less than 4364 grams. Round your answer to four decimal places.

Answer 1

Answer:$P(X<4364)=P(\frac{X-\mu}{\sigma}<\frac{4364-\mu}{\sigma})=P(Z<\frac{4364-3946}{607})=P(z<0.689)$And we can find this probability using the normal standard table or excel:

$P(z<0.689)=0.7546$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

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

$X \sim N(3946,\sqrt{368449}= 607)$  

Where $\mu=3946$ and $\sigma=607$

We are interested on this probability

$P(X<4364)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

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

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

$P(X<4364)=P(\frac{X-\mu}{\sigma}<\frac{4364-\mu}{\sigma})=P(Z<\frac{4364-3946}{607})=P(z<0.689)$And we can find this probability using the normal standard table or excel:

$P(z<0.689)=0.7546$