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QuicVO · Answer 1 · 2021-02-15T07:01:59+0000

Answer:

z=-0.920<(a-76)/(11)

And if we solve for a we got

a=76 -0.920*11=65.88

So the value of height that separates the bottom 17.88% of data from the top 82.12% is 65.88.

And for the other value we can use z = 0.920 since the distribution is symmetric, and we got:

a=76 +0.920*11=86.12

So the two values for this case who accumulate 64.24% of the area are (65.88; 86.12)

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:

$X \sim N(76,11)$

Where
$\mu=76$ and
$\sigma=11$

And we want 0.6424 of the area between two values. So then we have on the tails (1-0.6424)/2 = 0.1788 of the area.

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

P(X>a)=0.8212 (a)

P(X<a)=0.1788 (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 0.1788 of the area on the left and 0.8212 of the area on the right it's z=-0.920. On this case P(Z<-0.920)=0.1788 and P(z>-0.920)=0.8212

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

$P(X<a)=P((X-\mu)/(\sigma)<(a-\mu)/(\sigma))=0.1788$

$P(z<(a-\mu)/(\sigma))=0.1788$

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