# A population of values has a normal distribution with \mu=154.5 and \sigma=96.1. You intend to draw a random sample of size n=134. Find the probability that a sample of size n=134 is randomly selected with a mean greater than 167. P(M > 167) =? Write your answers as numbers accurate to 4 decimal places.

Question
Random variables
A population of values has a normal distribution with $$\displaystyle\mu={154.5}$$ and $$\displaystyle\sigma={96.1}$$. You intend to draw a random sample of size $$\displaystyle{n}={134}$$.
Find the probability that a sample of size $$\displaystyle{n}={134}$$ is randomly selected with a mean greater than 167.
$$\displaystyle{P}{\left({M}{>}{167}\right)}=$$?

2020-11-10
$$\displaystyle{X}_{{{1}}},{X}_{{{2}}},\ldots{X}_{{{134}}}$$ be a random sample from Normal distribution with $$\displaystyle\mu={154.5}$$ and $$\displaystyle\sigma={96.1}$$.
Then the sample mean
$$\displaystyle{M}={\frac{{{1}}}{{{134}}}}{\sum_{{{i}={1}}}^{{{134}}}}{X}_{{{i}}}$$ follows Normal distribution with mean 154.5 and standard deviation $$\displaystyle{\frac{{{96.1}}}{{\sqrt{{{134}}}}}}$$.
Hence, the probability that sample mean greater than 167.
$$\displaystyle{P}{\left({M}{>}{167}\right)}$$
$$\displaystyle={1}-{P}{\left({M}\leq{167}\right)}$$
$$\displaystyle={1}-{P}{\left({\frac{{{M}-{154.5}}}{{\frac{{96.1}}{\sqrt{{{134}}}}}}}\leq{\frac{{{167}-{154.5}}}{{\frac{{96.1}}{\sqrt{{{134}}}}}}}\right)}$$
$$\displaystyle={1}-{P}{\left({Z}\leq{1.506}\right)},{Z}={\left({\frac{{{M}-{154.5}}}{{\frac{{96.1}}{\sqrt{{{134}}}}}}}\right)}$$ follows Normal (0,1)
$$\displaystyle={1}-\phi{\left({1.506}\right)},\phi{\left({1.506}\right)}$$ calculated from Normal distribution table.
$$\displaystyle={1}-{0.934}={0.066}$$
Therefore, the probability that a sample of size $$\displaystyle{n}={134}$$ is randomly selected with a mean greater than 167 is 0.066.

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