Question

A population of values has a normal distribution with \mu=239.5 and \sigma=32.7.

Random variables
ANSWERED
asked 2020-10-25

A population of values has a normal distribution with \(\displaystyle\mu={239.5}\) and \(\displaystyle\sigma={32.7}\). You intend to draw a random sample of size \(\displaystyle{n}={139}\).
Find the probability that a sample of size \(n=139\) is randomly selected with a mean greater than 235.9.
\(\displaystyle{P}{\left({M}{>}{235.9}\right)}=\)?
Write your answers as numbers accurate to 4 decimal places.

Answers (1)

2020-10-26

From the provided information,
Mean \(\displaystyle{\left(\mu\right)}={239.5}\)
Standard deviation \(\displaystyle{\left(\sigma\right)}={32.7}\)
Sample size \(\displaystyle{\left({n}\right)}={139}\)
Let X be a random variable which represents the value score.
\(\displaystyle{X}\sim{N}{\left({239.5},{32.7}\right)}\)
The required probability that a sample of size \(\displaystyle{n}={139}\) is randomly selected with a mean greater than 235.9 can be obtained as:
\(\displaystyle{P}{\left({M}{>}{235.9}\right)}={P}{\left({\frac{{{M}-\mu}}{{{\frac{{\sigma}}{{\sqrt{{{n}}}}}}}}}{>}{\frac{{{235.9}-{239.5}}}{{{\frac{{{32.7}}}{{\sqrt{{{139}}}}}}}}}\right)}\)
\(\displaystyle={P}{\left({Z}\succ{1.298}\right)}\)
\(\displaystyle={1}-{P}{\left({Z}{<}-{1.298}\right)}\)
\(\displaystyle={1}-{0.0971}={0.9029}\) (Using standard normal table)
Hence, the required probability is 0.9029.

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