Question 1193652: Suppose that The lengths of pregnancies are normally distributed with a mean of 271 days and a standard deviation of 10 days. how do I find the percentage a pregnancy that are longer than 250 4 days for my immediate computations used for or more decimal places give my final answer to two decimal places.
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
I'm assuming you meant to write 250.4 instead of "250 4" with a space between the 250 and 4.
Given info:
mu = 271 = population mean of the length of pregnancies in days
sigma = 10 = population standard deviation of the length of pregnancies in days
Convert x = 250.4 to its corresponding z score
z = (x-mu)/sigma
z = (250.4-271)/10
z = -2.06
Now you can use a calculator like this one
https://onlinestatbook.com/2/calculators/normal_dist.html
to find that P(Z > -2.06) = 0.9803 approximately which converts to 98.03%
Or you can use a Z table like shown here
https://www.ztable.net/
and the table says that P(Z < -2.06) = 0.01970
Therefore,
P(Z > -2.06) = 1 - P(Z < -2.06)
P(Z > -2.06) = 1 - 0.01970
P(Z > -2.06) = 0.9803
This then converts back to
P(X > 250.4) = 0.9803
To show that about 98.03% of all pregnancies are longer than 250.4 days.
If you prefer to use a TI calculator, then you'll hit the button labeled "2ND" and then hit the "VARS" key to bring up the stats function menu. Scroll down to "normalcdf".
You can type in something like
normalcdf(-2.06, 99)
The 99 is to set up some upper boundary that is fairly large
Side note: With the calculator options, technically we don't need to convert to a z score because both calculator options allow us to change the mu and sigma values. Though it's often standard practice to translate to a standard normal z score and use mu = 0 and sigma = 1.
Answer: Approximately 98.03%
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