 # A population of values has a normal distribution with mean 18.6 and standard deviation 57. If a random sample of size n=25 is selected, Find the probability that a sample of size n=25 is randomly selected with a mean greater than 17.5 round your answer to four decimal places. P=? BolkowN 2020-11-08 Answered
A population of values has a normal distribution with mean 18.6 and standard deviation 57. If a random sample of size $n=25$ is selected,
Find the probability that a sample of size $n=25$ is randomly selected with a mean greater than 17.5 round your answer to four decimal places.
P=?
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A population of values has a normal distribution with mean $18.6$ and standard deviation $57$, sample size is $n=25$. The value of $x$ is $17.5$. The $z$-score is,
$z=\frac{x-\mu }{\frac{\sigma }{\sqrt{n}}}$
$=\frac{17.5-18.6}{\frac{57}{\sqrt{25}}}$
$=\frac{-1.1}{11.4}=-0.0965$
The area to the right of $z=-0.0965$ under the standard normal curve is $P\left(z>-0.0965\right)=1-P\left(z<-0.0965\right)$
The probability of $z$ less than $–0.0965$ can be obtained using the excel formula “$=NORM.S.DIST\left(–0.0965,TRUE\right)$”. The probability value is $0.4616$.
The probability that a sample of size $n=25$ is randomly selected with a mean greater than $17.5$ is,
$P\left(z\succ 0.0965\right)=1-P\left(z<-0.0965\right)$
$1-0.4616=0.5384$
Thus, the probability that a sample of size $n=25$ is randomly selected with a mean greater than $17.5$ is $0.5384.$

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