Question 1207570
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The lengths of pregnancies in a small rural village are normally distributed 
with a mean of 266 days and a standard deviation of 17 days.
What percentage of pregnancies last beyond 252 days?
P(X > 252 days) =
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<pre>
A normal distribution curve is a bell shaped curve.

The total area under each such curve is 1.

The percentage of pregnancies that last beyond 252 days, 
is the area under this normal curve on the right of the raw mark X= 252 days, 


You can solve the problem using a regular calculator like TI-83/84.
For the given raw mark, such calculators provide the area on the LEFT of the raw mark,
Therefore, you should take the COPMPLEMENT to it.

So, for the percentage of pregnancies that last beyond 252 days, 

                       z1    z2   mu   SD    <<<---===  formatting pattern
     P = 1 - normalcdf(252, 9999, 266, 17).


Here nomalcdf stands for normal cumulative distribution function.


For normalcdf(252, 9999, 266, 17), your calculator will give you the value  P' = 0.2051.


Thus, your final answer is the complement to it  P = 1 - 0.2051 = 0.7949.    <U>ANSWER</U>



Another way to solve the problem is to use an online calculator for normal distribution.

Go to web-site 

    https://onlinestatbook.com/2/calculators/normal_dist.html

and use free of charge online calculator there, which is intended for the same purposes.

This online calculator has perfect and intuitively clear interface, so even a beginner student 
can use it without any difficulties.  Surely, this online calculator will get you the same answer.
The calculator shows the area of the interest, so after using it, you will understand the meaning 
of the procedure in full.

So, choose the option "Above" and get the same final probability P = 0.7949 without any trouble 
about taking the complementary probability.
</pre>

Solved.