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

CheemnCatelvew 2020-10-18 Answered
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 single randomly selected value is greater than 235.9.
\(\displaystyle{P}{\left({X}{>}{235.9}\right)}=\)?
Write your answers as numbers accurate to 4 decimal places.

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Expert Answer

izboknil3
Answered 2020-10-19 Author has 15890 answers

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 single randomly selected value is greater than 235.9 can be obtained as:
\(\displaystyle{P}{\left({X}{>}{235.9}\right)}={P}{\left({\frac{{{x}-\mu}}{{\sigma}}}{>}{\frac{{{235.9}-{239.5}}}{{{32.7}}}}\right)}\)
\(\displaystyle={P}{\left({Z}\succ{0.110}\right)}\)
\(\displaystyle={1}-{P}{\left({Z}{<}-{0.110}\right)}\)
\(\displaystyle={1}-{0.4562}={0.5438}\) (Using standard normal table)
Thus, the required probability is 0.5438.

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