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Suppose that the mean score on a creativity test is 16 and the standard deviation is 4. You are told that the distribution is normal. Using the approximations for normal curves, how many people would get a score between 12 and 20?

User Mohsenme
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1 Answer

4 votes

Answer:


P(12<X<20)=P((12-\mu)/(\sigma)<(X-\mu)/(\sigma)<(20-\mu)/(\sigma))=P((12-16)/(4)<Z<(20-16)/(4))=P(-1<Z<1)

And we can find this probability on this way:


P(-1<Z<1)=P(Z<1)-P(Z<-1) =0.84113-0.15865 =0.6827

We expect around 68.27% between the two scores provided.

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(16,4)

Where
\mu=16 and
\sigma=4

We are interested on this probability


P(12<X<20)

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


z=(x-\mu)/(\sigma)

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


P(12<X<20)=P((12-\mu)/(\sigma)<(X-\mu)/(\sigma)<(20-\mu)/(\sigma))=P((12-16)/(4)<Z<(20-16)/(4))=P(-1<Z<1)

And we can find this probability on this way:


P(-1<Z<1)=P(Z<1)-P(Z<-1) =0.84113-0.15865 =0.6827

We expect around 68.27% between the two scores provided.

User Sridhar DD
by
8.9k points
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