Question

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

Random variables
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asked 2021-02-21

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 size \({n}={42}\).
Find the probability that a single randomly selected value is between 221.6 and 229.1.
\(\displaystyle{P}{\left({221.6}{<}{X}{<}{229.1}\right)}=\)?
Write your answers as numbers accurate to 4 decimal places.

Expert Answers (1)

2021-02-22

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.
\(\displaystyle{X}\sim{N}{\left({221.6},{44.1}\right)}\)
Step 2
The probability that a single randomly selected value is between 221.6 and 229.1 is \(\displaystyle{P}{\left({221.6}{<}{X}{<}{229.1}\right)}=\)
\(\displaystyle{P}{\left({221.6}{<}{X}{<}{229.1}\right)}={P}{\left({\frac{{{221.6}-{221.6}}}{{\frac{{44.1}}{\sqrt{{{1}}}}}}}{<}{\frac{{{X}-\mu}}{{\frac{\sigma}{\sqrt{{{n}}}}}}}{<}{\frac{{{229.1}-{221.6}}}{{\frac{{44.1}}{\sqrt{{{1}}}}}}}\right)}\)
\(\displaystyle{P}{\left({221.6}{<}{X}{<}{229.1}\right)}={P}{\left({0}{<}{Z}{<}{0.1701}\right)}\)
\(\displaystyle{P}{\left({221.6}{<}{X}{<}{229.1}\right)}={P}{\left({Z}{<}{0.170}\right)}-{P}{\left({Z}{<}{0}\right)}\)
\(\displaystyle{P}{\left({221.6}{<}{X}{<}{229.1}\right)}={0.5675}-{0.5}={0.0675}\)
(From excel using formula = NORM.S. DIST (0.170,TRUE))
(From excel using formula = NORM.S. DIST (0,TRUE))
Hence, the probability that a single randomly selected value is between 221.6 and 229.1 is 0.0675.

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