Question

Based on data from a large study of healthy infants in six countries, the World Health Organization produced growth charts that are part of every pediatrician’s toolkit for monitoring a child’s overall health. According to the WHO report, girls who are one month old have a mean head circumference of 36.6 centimeters with a standard deviation of 1.2 centimeters. As with most body measurements, head circumference has a normal probability distribution. Medscape defines microcephaly (small head syndrome) as a head circumference that is more than two standard deviations below the mean. What is the probability that a one-month old girl will be categorized as having microcephaly? Group of answer choices 68% 95% 5% 2.5%

Answer 1

Answer:

We are interested on this probability

$P(X<\mu -2\sigma = 36.6 -2*1.2 =34.2)$

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:

$P(X<34.2)=P(\frac{X-\mu}{\sigma}<\frac{34.2-\mu}{\sigma})=P(Z<\frac{34.2-36.6}{1.2})=P(z<-2)$

And we can find this probability using the normal standard table or excel and we got:

$P(z<-2)=0.025$

And the best answer for this case would be:

2.5%

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 head circumference of a population, and for this case we know the distribution for X is given by:

$X \sim N(36.6,1.2)$  

Where $\mu=36.6$ and $\sigma=1.2$

We are interested on this probability

$P(X<\mu -2\sigma = 36.6 -2*1.2 =34.2)$

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:

$P(X<34.2)=P(\frac{X-\mu}{\sigma}<\frac{34.2-\mu}{\sigma})=P(Z<\frac{34.2-36.6}{1.2})=P(z<-2)$

And we can find this probability using the normal standard table or excel and we got:

$P(z<-2)=0.025$

And the best answer for this case would be:

2.5%