An elevator has a placard stating that the maximum capacity is 1376 lb—8 passengers. So, 8 adult male passengers can have a mean weight of up to 1376/8=172 pounds. If the elevator is loaded with 8 adult male passengers, find the probability that it is overloaded because they have a mean weight greater than 172 lb. (Assume that weights of males are normally distributed with a mean of 178 lb and a standard deviation of 27 lb.) Does this elevator appear to be safe?
The probability the elevator is overloaded is
.
Given : Mean "(\\mu)=174" , standard deviation "(\\sigma)=28" , sample size "(n)=8"
Formula: "z=\\frac{\\bar{x}-\\mu}{\\frac{\\sigma}{\\sqrt{n}}}"
Find: "P\\text{( sample mean greater than 172)}=P(\\bar{x}>172)=?"
"\\begin{aligned}\n\n&\\Rightarrow P\\left(\\frac{\\bar{x}-\\mu}{\\frac{\\sigma}{\\sqrt{n}}}>\\frac{172-174}{\\frac{28}{\\sqrt{8}}}\\right) \\\\\n\n&\\Rightarrow P(z>-0.2020)\n\n\\end{aligned}"
Note: "P(z>a)=1-P(z<a)"
"\\Rightarrow 1-P(z<-0.2020)"
Refer Standard normal table/Z-table to find the probability or use excel formula "=NORM.S.DIST(-0.2020, TRUE)" to find the probability.
"\\begin{aligned}\n\n&\\Rightarrow 1-0.4200 \\\\\n\n&\\Rightarrow {0 . 5 8 0 0}\n\n\\end{aligned}"
The probability the elevator is overloaded is 0.5800
No, there is a good chance that 8 randomly selected people will exceed the elevator capacity.
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