Question 1199247
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n = 42 = sample size


Assumption is that male weights are normally distributed with these parameters:
mu = 175.9 = population mean
sigma = 38.4 = population standard deviation


xbar = sample mean


The xbar distribution is the set of xbar values that randomly scatter about, but should be close to the value of mu = 175.9
The spread of this xbar distribution is: sigma/sqrt(n) = 38.4/sqrt(42) = 5.9252486 approximately.


In short, the xbar distribution we're after has these properties:
center = 175.9
spread = 5.9252486 approximately


Let's compute the z score for xbar = 169 pounds.
z = (xbar - center)/(spread)
z = (169 - 175.9)/(5.9252486)
z = -1.1645081018204
z = -1.16
This result is approximate.


The task of finding P(xbar > 169) is roughly equivalent to P(z > -1.16)


Now use a calculator such as 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 > -1.16) = 0.877
approximately.
Using a Z table is another option you can take. 


There's about an 87.7% chance of the average passenger being over 169 pounds. 
This by extension leads to about an 87.7% chance of the plane being overloaded.


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Answer: 


The probability of an overloaded aircraft is approximately 0.877, so the pilot should take action to correct this.
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