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".

2) Solution to the problem

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

$X \sim N(150.7,10.2)$

Where $\mu=150.7$ and $\sigma=10.2$

We want to find a value a, such that we satisfy this condition:

$P(X>a)=0.65$ (a)

$P(X$ (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 0.35 of the area on the left and 0.65 of the area on the right it's z=-0.385. On this case P(Z<-0.385)=0.35 and P(Z>-0.385)=0.65

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

$P(X$

$P(z$

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

$Z=-0.385$

And if we solve for a we got

$a=150.7 -0.385*10.2=146.773$

So the value of height that separates the bottom 35% of data from the top 65% (Or the 35 percentile) is 146.7.

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3 years ago

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