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.

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
asked 2020-10-26
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>?
Write your answers as numbers accurate to 4 decimal places.

Answers (1)

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.
0

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