A population of values has a normal distribution with \mu = 200 and \sigma = 31.9. You intend to draw a random sample of size n = 11. Find the probability that a sample of size n = 11 is randomly selected with a mean less than 226.9. P(M<226.9)=? Write your answers as numbers accurate to 4 decimal places.

A population of values has a normal distribution with \mu = 200 and \sigma = 31.9. You intend to draw a random sample of size n = 11. Find the probability that a sample of size n = 11 is randomly selected with a mean less than 226.9. P(M<226.9)=? Write your answers as numbers accurate to 4 decimal places.

Question
Random variables
asked 2021-01-16
A population of values has a normal distribution with \(\displaystyle\mu={200}\) and \(\displaystyle\sigma={31.9}\). You intend to draw a random sample of size \(\displaystyle{n}={11}\).
Find the probability that a sample of size \(\displaystyle{n}={11}\) is randomly selected with a mean less than 226.9.
\(\displaystyle{P}{\left({M}{<}{226.9}\right)}=\)</span>?
Write your answers as numbers accurate to 4 decimal places.

Answers (1)

2021-01-17
The probability that a sample of size \(\displaystyle{n}={11}\) is randomly selected with a mean less than 226.9 is,
\(\displaystyle{P}{\left({M}{<}{226.9}\right)}={P}{\left({\frac{{{M}-\mu}}{{{\left({\frac{{\sigma}}{{\sqrt{{{n}}}}}}\right)}}}}{<}{\frac{{{226.9}-{200}}}{{{\left({\frac{{{31.9}}}{{\sqrt{{{11}}}}}}\right)}}}}\right)}\)</span>
\(\displaystyle={P}{\left({z}{<}{\frac{{{26.9}}}{{{9.6182}}}}\right)}\)</span>
\(\displaystyle={P}{\left({z}{<}{2.797}\right)}\)</span>
The probability of z less than 2.797 can be obtained using the excel formula “=NORM.S.DIST(2.797,TRUE)”. The probability value is 0.9974.
The required probability value is,
\(\displaystyle{P}{\left({M}{<}{226.9}\right)}={P}{\left({z}{<}{2.797}\right)}={0.9974}\)</span>
Thus, the probability that a sample of size \(\displaystyle{n}={11}\) is randomly selected with a mean less than 226.9 is 0.9974.
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