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

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Random variables 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.
$$\displaystyle{P}{\left({98.5}{<}\overline{{{X}}}{<}{100.7}\right)}=$$</span>?
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:
$$\displaystyle{P}{\left({98.5}{<}\overline{{{X}}}{<}{100.7}\right)}={P}{\left({\frac{{{98.5}-{99.6}}}{{{\frac{{{35.1}}}{{\sqrt{{{84}}}}}}}}}{<}{\frac{{{x}-\mu}}{{{\frac{{\sigma}}{{\sqrt{{{n}}}}}}}}}{<}{\frac{{{100.7}-{99.6}}}{{{\frac{{{35.1}}}{{\sqrt{{{84}}}}}}}}}\right)}$$</span>
$$\displaystyle={P}{\left(-{0.287}{<}{Z}{<}{0.287}\right)}$$</span>
$$\displaystyle={P}{\left({Z}{<}{0.287}\right)}-{P}{\left({Z}{<}-{0.287}\right)}$$</span>
$$\displaystyle={P}{\left({Z}{<}{0.287}\right)}-{\left[{1}-{P}{\left({z}{<}{0.287}\right)}\right]}$$</span>
$$\displaystyle={2}{P}{\left({Z}{<}{0.287}\right)}-{1}$$</span>
$$\displaystyle={2}{\left({0.6129}\right)}-{1}={0.2258}$$ (Using standard normal table)
Thus, the required probability is 0.2258.

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