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# A population of values has a normal distribution with \mu=77 and \sigma=32.2.

Random variables
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asked 2020-12-22

A population of values has a normal distribution with $$\displaystyle\mu={77}$$ and $$\displaystyle\sigma={32.2}$$. You intend to draw a random sample of size $$\displaystyle{n}={15}$$
Find the probability that a sample of size $$\displaystyle{n}={15}$$ is randomly selected with a mean between 59.5 and 98.6. $$\displaystyle{P}{\left({59.5}{<}\overline{{{X}}}{<}{98.6}\right)}=$$?
Write your answers as numbers accurate to 4 decimal places.

## Expert Answers (1)

2020-12-23

From the given information, $$\displaystyle\mu={77}$$, $$\displaystyle\sigma={32.2}$$ and the sample size=15.
Consider,
$$\displaystyle{P}{\left({59.5}{<}\overline{{{X}}}{<}{98.6}\right)}={P}{\left({\frac{{{59.5}-\mu}}{{{\left({\frac{{\sigma}}{{\sqrt{{{n}}}}}}\right)}}}}{<}{\frac{{\overline{{{X}}}-\mu}}{{{\left({\frac{{\sigma}}{{\sqrt{{{n}}}}}}\right)}}}}{<}{\frac{{{98.6}-\mu}}{{{\left({\frac{{\sigma}}{{\sqrt{{{n}}}}}}\right)}}}}\right)}$$
$$\displaystyle={P}{\left({\frac{{{59.5}-\mu}}{{{\left({\frac{{{32.2}}}{{\sqrt{{{15}}}}}}\right)}}}}{<}{z}{<}{\frac{{{98.6}-{77}}}{{{\left({\frac{{{32.2}}}{{\sqrt{{{15}}}}}}\right)}}}}\right)}$$
$$\displaystyle={P}{\left(-{2.10488}{<}{z}{<}{2.59803}\right)}$$
$$\displaystyle={P}{\left({z}{<}{2.59803}\right)}-{P}{\left({z}{<}-{2.10488}\right)}$$
$$=0.748828-0.2934=0.4554 \begin{bmatrix}From\ the\ Excel\ function \\=NORM.DIST(0.6770807,0,1,TRUE) \\=NORM.DIST(-0.54348,0,1,TRUE) \end{bmatrix}$$
Thus, the probability that a sample of size $$\displaystyle{n}={15}$$ is randomly selected with a mean between 59.5 and 98.6 is 0.9777.

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