A population of values has a normal distribution with \mu = 116.5 and \sigma = 63.7. You intend to draw a random sample of size n = 244. Find the probability that a sample of size n = 244 is randomly selected with a mean between 104.7 and 112.8. P(104.7 < M < 112.8)=? Write your answers as numbers accurate to 4 decimal places.

Question
Random variables
asked 2020-11-29
A population of values has a normal distribution with \(\displaystyle\mu={116.5}\) and \(\displaystyle\sigma={63.7}\). You intend to draw a random sample of size \(\displaystyle{n}={244}\).
Find the probability that a sample of size \(\displaystyle{n}={244}\) is randomly selected with a mean between 104.7 and 112.8.
\(\displaystyle{P}{\left({104.7}{<}{M}{<}{112.8}\right)}=\)</span>?
Write your answers as numbers accurate to 4 decimal places.

Answers (1)

2020-11-30
A population of values has a normal distribution with \(\displaystyle\mu={116.5}\) and \(\displaystyle\sigma={63.7}\).
A random sample of size \(\displaystyle{\left({n}\right)}={244}\) is drawn.
We need to find the probability that a sample of size \(\displaystyle{n}={244}\) is randomly selected with a mean(M) between 104.7 and 112.8.
\(\displaystyle{M}\sim{N}{\left({116.5},{\frac{{{63.7}^{{{2}}}}}{{{n}}}}\right)}\) and \(\displaystyle{n}={244}\)
So, \(\displaystyle{Z}={\frac{{\sqrt{{{n}}}{\left({M}-{116.5}\right)}}}{{{63.7}}}}\sim{N}{\left({0},{1}\right)}\)
Thus, \(\displaystyle{P}{\left[{104.7}{<}{M}{<}{112.8}\right]}\)</span>
\(\displaystyle={P}{\left[{\frac{{\sqrt{{{n}}}{\left({104.7}-{116.5}\right)}}}{{{63.7}}}}{<}{\frac{{\sqrt{{{n}}}{\left({M}-{116.5}\right)}}}{{{63.7}}}}{<}{\frac{{\sqrt{{{n}}}{\left({112.8}-{116.5}\right)}}}{{{63.7}}}}\right]}\)</span>
\(\displaystyle={P}{\left[-{2.893}{<}{Z}{<}-{0.9073}\right]}\)</span>
\(\displaystyle={P}{\left[{Z}\leq-{0.9073}\right]}-{P}{\left[{Z}\leq-{2.8934}\right]}\)
\(\displaystyle={0.1821}-{0.0019}={0.1802}\)
So, the probability that a sample of size \(\displaystyle{n}={244}\) is randomly selected with a mean between 104.7 and 112.8 is 0.1802.
0

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