Question

Suppose the ice Q scores in one region are normally distributed within a standard deviation of 13. Suppose also that exactly 51% of individuals from this region have IQ scores of greater than 100 and that 49% do not. What is the mean IQ score for this region?

Answer 1

Answer:

$z=-0.025<\frac{100-\mu}{12}$

And if we solve for $\mu$ we got

$\mu=100 +0.025*13=100.325$

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

$X \sim N(\mu,13)$  

Where $\mu=?$ and $\sigma=13$

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

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

For this case we know these conditions:

$P(X>100)=0.51$   (a)

$P(X<100)=0.49$   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.49 of the area on the left and 0.51 of the area on the right it's z=-0.025. On this case P(Z<-0.025)=0.49 and P(z>-0.025)=0.51

If we use condition (b) from previous we have this:

$P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.49$  

$P(z<\frac{a-\mu}{\sigma})=0.49$

But we know which value of z satisfy the previous equation so then we can do this:

$z=-0.025<\frac{100-\mu}{12}$

And if we solve for $\mu$ we got

$\mu=100 +0.025*13=100.325$