Question

The duration (in days) of human pregnancies follows approximately the N(266,16) distribution. How many days would a human pregnancy need to last to be among the top 10% of all durations?

Answer 1

`a=266 +1.28*16=286.48`

So the value of height that separates the bottom 90% of data from the top 10% is 286.48

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 duration in days of human pregnancis of a population, and for this case we know the distribution for X is given by:

`X \sim N(266,16)`

Where
`\mu =266` and
`\sigma=16`

The z score given by this formula:

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

We want the upper 10% of the values. So for this part we want to find a value a, such that we satisfy this condition:

`P(X>a)=0.1` (a)

`P(X<a)=0.9` (b)

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

As we can see on the figure attached the z value that satisfy the condition with 0.9 of the area on the left and 0.1 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.9 and P(z>1.28)=0.1

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

`P(X<a)=P((X-\mu)/(\sigma)<(a-\mu)/(\sigma))=0.9`

`P(z<(a-\mu)/(\sigma))=0.9`

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

`z=1.28<(a-266)/(16)`

And if we solve for a we got

`a=266 +1.28*16=286.48`

So the value of height that separates the bottom 90% of data from the top 10% is 286.48