Question

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

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asked 2021-02-11
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.

Answers (1)

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.
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