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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?

User Han He
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1 Answer

1 vote

Answer:


z=-0.025<(100-\mu)/(12)

And if we solve for
\mu we got


\mu=100 +0.025*13=100.325

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=(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((X-\mu)/(\sigma)<(a-\mu)/(\sigma))=0.49


P(z<(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<(100-\mu)/(12)

And if we solve for
\mu we got


\mu=100 +0.025*13=100.325

User Mamesaye
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4.2k points