Question

A population of values has a normal distribution with \mu=99.6 and \sigma=35.1. You intend to draw a random

Random variables
ANSWERED
asked 2021-01-05

A population of values has a normal distribution with \(\displaystyle\mu={99.6}\) and \(\displaystyle\sigma={35.1}\). You intend to draw a random sample of size \(\displaystyle{n}={84}\).
Find the probability that a sample of size \(\displaystyle{n}={84}\) is randomly selected with a mean between 98.5 and 100.7.
\(\displaystyle{P}{\left({98.5}{<}\overline{{{X}}}{<}{100.7}\right)}=\)?
Write your answers as numbers accurate to 4 decimal places.

Answers (1)

2021-01-06

Step 1
From the provided information,
Mean \(\displaystyle{\left(\mu\right)}={99.6}\)
Standard deviation \(\displaystyle{\left(\sigma\right)}={35.1}\)
\(\displaystyle{X}\sim{N}{\left({99.6},{35.1}\right)}\)
Sample size \(\displaystyle{\left({n}\right)}={84}\)
Step 2
The required probability that a sample of size \(\displaystyle{n}={84}\) is randomly selected with a mean between 98.5 and 100.7 can be obtained as:
and 100.7 can be obtained as:
\(\displaystyle{P}{\left({98.5}{<}\overline{{{X}}}{<}{100.7}\right)}={P}{\left({\frac{{{98.5}-{99.6}}}{{{\frac{{{35.1}}}{{\sqrt{{{84}}}}}}}}}{<}{\frac{{{x}-\mu}}{{{\frac{{\sigma}}{{\sqrt{{{n}}}}}}}}}{<}{\frac{{{100.7}-{99.6}}}{{{\frac{{{35.1}}}{{\sqrt{{{84}}}}}}}}}\right)}\)
\(\displaystyle={P}{\left(-{0.287}{<}{Z}{<}{0.287}\right)}\)
\(\displaystyle={P}{\left({Z}{<}{0.287}\right)}-{P}{\left({Z}{<}-{0.287}\right)}\)
\(\displaystyle={P}{\left({Z}{<}{0.287}\right)}-{\left[{1}-{P}{\left({z}{<}{0.287}\right)}\right]}\)
\(\displaystyle={2}{P}{\left({Z}{<}{0.287}\right)}-{1}\)
\(\displaystyle={2}{\left({0.6129}\right)}-{1}={0.2258}\) (Using standard normal table)
Thus, the required probability is 0.2258.

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