A population of values has a normal distribution with \mu=221.6 and \sigma=44.1. You intend to draw a random sample of size n=42. Find the probability that a sample of size n=42 is randomly selected with a mean between 221.6 and 229.1. P(221.6 < \bar{X} < 229.1) =? Write your answers as numbers accurate to 4 decimal places.

Question
Random variables
A population of values has a normal distribution with $$\displaystyle\mu={221.6}$$ and $$\displaystyle\sigma={44.1}$$. You intend to draw a random sample of $$\displaystyle{s}{i}{z}{e}{n}={42}$$.
Find the probability that a sample of size $$\displaystyle{n}={42}$$ is randomly selected with a mean between 221.6 and 229.1.
$$\displaystyle{P}{\left({221.6}{<}\overline{{{X}}}{<}{229.1}\right)}=$$</span>?

2020-10-27
Step 1
Given information-
Population mean,$$\displaystyle\mu={221.6}$$
Population standard deviation, $$\displaystyle\sigma={44.1}$$
Sample size, $$\displaystyle{n}={42}$$
Let, X be the randomly selected value is approximately normally distributed.
$$\displaystyle{X}\sim{N}{\left({221.6},{44.1}\right)}$$
The probability that a sample of size n=42 is randomly selected with a mean between 221.6 and 229.1-
$$\displaystyle{P}{\left({221.6}{<}{X}{<}{229.1}\right)}={P}{\left({\frac{{{221.6}-{221.6}}}{{\frac{{44.1}}{\sqrt{{{42}}}}}}}{<}{\frac{{{X}-\mu}}{{\frac{\sigma}{\sqrt{{{n}}}}}}}{<}{\frac{{{229.1}-{221.6}}}{{\frac{{44.1}}{\sqrt{{{42}}}}}}}\right)}$$</span>
$$\displaystyle{P}{\left({221.6}{<}{X}{<}{229.1}\right)}={P}{\left({0}{<}{Z}{<}{1.102}\right)}$$</span>
$$\displaystyle{P}{\left({221.6}{<}{X}{<}{229.1}\right)}={P}{\left({Z}{<}{1.102}\right)}-{P}{\left({Z}{<}{0}\right)}$$</span>
$$\displaystyle{P}{\left({221.6}{<}{X}{<}{229.1}\right)}={0.8648}-{0.5}={0.3648}$$</span>
(From excel using formula = NORM.S. DIST (0.102,TRUE))
(From excel using formula = NORM.S. DIST (0,TRUE))
Hence, the probability that a sample of size $$\displaystyle{n}={42}$$ is randomly selected with a mean between 221.6 and 229.1 is 0.3648.

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