# A population of values has a normal distribution with \mu=116.3 and \sigma=27.5. You intend to draw a random sample of size n=249. Find the probabilit

Random variables
A population of values has a normal distribution with $$\displaystyle\mu={116.3}$$ and $$\displaystyle\sigma={27.5}$$. You intend to draw a random sample of size $$\displaystyle{n}={249}$$.
Find the probability that a sample of size $$\displaystyle{n}={249}$$ is a randomly selected with a mean greater than 117.3.
$$\displaystyle{P}{\left(\overline{{{X}}}{>}{117.3}\right)}=$$?
Write your answers as numbers accurate to 4 decimal places.

2021-02-12
$$\displaystyle\overline{{{X}}}={\frac{{{1}}}{{{n}}}}{\sum_{{{i}={1}}}^{{n}}}{X}_{{i}}=$$
Mean value.
Therefore, the mean of $$\displaystyle\overline{{{X}}}$$ is 116.3,
standard deviation is $$\displaystyle{\frac{{{27.5}}}{{\sqrt{{{n}}}}}}={\frac{{{27.5}}}{{\sqrt{{{249}}}}}}$$, as $$\displaystyle{n}={249}$$.
The probability of a mean greater than 117.3 is
$$\displaystyle{P}{\left[\overline{{{X}}}{>}{117.3}\right]}$$
$$\displaystyle={P}{\left[{\frac{{\overline{{{X}}}-{116.3}}}{{\frac{{27.5}}{\sqrt{{{249}}}}}}}{>}{\frac{{{117.3}-{116.3}}}{{\frac{{27.5}}{\sqrt{{{249}}}}}}}\right]}$$
$$\displaystyle={P}{\left[{Z}{>}{\frac{{\sqrt{{{249}}}}}{{{27.5}}}}\right]},{Z}={\frac{{\overline{{{X}}}-{116.3}}}{{\frac{{27.5}}{\sqrt{{{249}}}}}}}\sim$$ Normal(0,1)
$$\displaystyle={1}-{P}{\left[{Z}\leq{\frac{{\sqrt{{{249}}}}}{{{27.5}}}}\right]}$$
$$\displaystyle={1}-\phi{\left({0.574}\right)}$$
$$\displaystyle={1}-{0.717}={0.283}$$
The value of $$\displaystyle\phi{\left({0.574}\right)}$$ is taken from the Normal distribution table.
Therefore, the probability of the mean is greater than 117.3 is 0.283.