Question 1193652
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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
<a href = "https://onlinestatbook.com/2/calculators/normal_dist.html">https://onlinestatbook.com/2/calculators/normal_dist.html</a>
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
<a href = "https://www.ztable.net/">https://www.ztable.net/</a>
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: <font color=red>Approximately 98.03%</font>
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