Question

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

Random variables
ANSWERED
asked 2020-10-28

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 single randomly selected value is less than 226.9.
\(\displaystyle{P}{\left({X}{<}{226.9}\right)}=\)?
Write your answers as numbers accurate to 4 decimal places.

Answers (1)

2020-10-29

The probability that a single randomly selected value is less than 226.9 is,
\(\displaystyle{P}{\left({X}{<}{226.9}\right)}={P}{\left({\frac{{{X}-\mu}}{{\sigma}}}{<}{\frac{{{226.9}-{200}}}{{{31.9}}}}\right)}\)
\(\displaystyle={P}{\left({z}{<}{\frac{{{26.9}}}{{{31.9}}}}\right)}\)
\(\displaystyle={P}{\left({z}{<}{0.843}\right)}\)
The probability of z less than 0.843 can be obtained using the excel formula “=NORM.S.DIST(0.843,TRUE)”. The probability value is 0.8004.
The required probability value is,
\(\displaystyle{P}{\left({X}{<}{226.9}\right)}={P}{\left({z}{<}{0.843}\right)}={0.8004}\)
Thus, the probability that a single randomly selected value is less than 226.9 is 0.8004.

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