A population of values has a normal distribution with \mu=198.8 and \sigma=69.2. You intend to draw a random sample of size n=147.Find the probability that

A population of values has a normal distribution with $\mu =198.8$ and $\sigma =69.2$. You intend to draw a random sample of size $n=147$.
Find the probability that a sample of size $n=147$ is randomly selected with a mean between 184 and 205.1.
$P\left(184?

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The mean is 198.8, standard deviation is 69.2, and sample size is 147.
The probability that a sample of size $$\displaystyle{n}={147}$$ is randomly selected with a mean between 184 and 205.1 is,
$$\displaystyle{P}{\left({184}{<}{M}{<}{205.1}\right)}={P}{\left({\frac{{{184}-{198.8}}}{{{\left({\frac{{{69.2}}}{{\sqrt{{{147}}}}}}\right)}}}}{<}{\frac{{{M}-\mu}}{{{\left({\frac{{\sigma}}{{\sqrt{{{n}}}}}}\right)}}}}{<}{\frac{{{205.1}-{198.8}}}{{{\left({\frac{{{69.2}}}{{\sqrt{{{147}}}}}}\right)}}}}\right)}$$
$$\displaystyle={P}{\left({\frac{{-{14.8}}}{{{5.70775}}}}{<}{z}{<}{\frac{{{6.3}}}{{{5.7075}}}}\right)}$$
$$\displaystyle={P}{\left(-{2.593}{<}{z}{<}{1.104}\right)}$$
$$\displaystyle={P}{\left({z}{<}{1.104}\right)}-{P}{\left({z}{<}-{2.593}\right)}$$
The probability of z less than 1.104 can be obtained using the excel formula “=NORM.S.DIST(1.104,TRUE)”. The probability value is 0.8652.
The probability of z less than –2.593 can be obtained using the excel formula “=NORM.S.DIST(–2.593,TRUE)”. The probability value is 0.0048.
The required probability value is,
$$\displaystyle{P}{\left({184}{<}{M}{<}{205.1}\right)}={P}{\left({z}{<}{1.104}\right)}-{P}{\left({z}{<}-{2.593}\right)}$$
$$\displaystyle={0.8652}−{0.0048}={0.8604}$$
Thus, the probability that a sample of size $$\displaystyle{n}={147}$$ is randomly selected with a mean between 184 and 205.1 is 0.8604.

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