Question 1207570: 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) =
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52781)   (Show Source):
You can put this solution on YOUR website! .
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) =
~~~~~~~~~~~~~~~~~~~~~~
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. ANSWER
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.
Solved.
Answer by math_tutor2020(3817)  (Show Source):
You can put this solution on YOUR website!
You can use a specialized Z calculator as the other tutor mentions.
I'll use a Z table.
First we need to compute the z score.
z = (x - mu)/sigma
z = (252 - 266)/17
z = -0.82 approximately
I'm rounding to two decimal places because it is standard for many Z tables.
Such tables can be found in the back of your stats textbook.
If you don't have your textbook with you, then you can use an online resource such as this
https://www.ztable.net/
Let me know if you have questions on how to read the table.
Using such a table will determine that
P(Z < -0.82) = 0.20611
and then that leads to this
P(Z > -0.82) = 1-P(Z < -0.82)
P(Z > -0.82) = 1-0.20611
P(Z > -0.82) = 0.79389
Each decimal value mentioned is approximate.
That translates back to P(X > 252) = 0.79389 approximately when the mean is mu = 266 and the standard deviation is sigma = 17.
The percentage of pregnancies that last beyond 252 days is approximately 79.389% depending how you round it.
This accuracy can be improved through the use of a specialized Z calculator.
Some teachers will allow specialized Z calculators; while other teachers will only allow a stats table and standard calculator.
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