# A population of values has a normal distribution with \mu = 133.5 and \sigma = 5.2. You intend

he298c 2021-02-05 Answered

A population of values has a normal distribution with $\mu =133.5$ and $\sigma =5.2$. You intend to draw a random sample of size $n=230$.
Find the probability that a sample of size $n=230$ is randomly selected with a mean between 133.6 and 134.1.
$P\left(133.6<\stackrel{―}{x}<134.1\right)=$?
Write your answers as numbers accurate to 4 decimal places.

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## Expert Answer

delilnaT
Answered 2021-02-06 Author has 94 answers

Given Information:
$\mu =133.5$ and $\sigma =5.2$
$n=230$
To find the probability that a sample of size $n=230$ is randomly selected with a mean between 133.6 and 134.1:
Based on the concept of Central limit theorem, if a random sample of size 'n' is drawn from a population with mean $\mu$ and standard deviation $\sigma$, the sampling distribution of sample mean $\stackrel{―}{x}$ is approximately normally distributed with mean ${\mu }_{\stackrel{―}{x}}=\mu$ and standard deviation ${\sigma }_{\stackrel{―}{x}}=\frac{\sigma }{\sqrt{n}}$.
Mean of sample mean = ${\mu }_{\stackrel{―}{x}}=\mu =133.5$
Standard deviation ${\sigma }_{\stackrel{―}{x}}=\frac{\sigma }{\sqrt{n}}=\frac{5.2}{\sqrt{230}}=0.343$
Required probability can be obtained as follows:
$P\left(133.6<\stackrel{―}{x}<134.1\right)=P\left(\stackrel{―}{x}<134.1\right)-P\left(\stackrel{―}{x}<133.6\right)$
$=P\left(\frac{\stackrel{―}{x}-\mu }{\sigma }<\frac{134.1-133.5}{0.343}\right)-P\left(\frac{\stackrel{―}{x}-\mu }{\sigma }<\frac{133.6-133.5}{0.343}\right)$
$=P\left(Z<1.75\right)-P\left(Z<0.29\right)$

Therefore, probability that a sample of size $n=230$ is randomly selected with a mean between 133.6 and 134.1 is 0.3459

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Jeffrey Jordon
Answered 2021-11-14 Author has 2070 answers

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