Question

A population of values has a normal distribution with \mu=198.8 and \sigma=69.2. You intend to draw a random sample of size n=147.Find the probability that a single randomly selected value is between 184 and 205.1.P(184 < X < 205.1) =?Write your answers as numbers accurate to 4 decimal places.

Random variables
ANSWERED
asked 2020-10-26

A population of values has a normal distribution with \(\displaystyle\mu={198.8}\) and \(\displaystyle\sigma={69.2}\). You intend to draw a random sample of size \(\displaystyle{n}={147}\).
Find the probability that a single randomly selected value is between 184 and 205.1.
\(\displaystyle{P}{\left({184}{<}{X}{<}{205.1}\right)}=\)?
Write your answers as numbers accurate to 4 decimal places.

Expert Answers (1)

2020-10-27

The mean is 198.8, standard deviation is 69.2, and sample size is 147.
The probability that a single randomly selected value is between 184 and 205.1 is,
\(\displaystyle{P}{\left({184}{<}{X}{<}{205.1}\right)}={P}{\left({\frac{{{184}-{198.8}}}{{{69.2}}}}{<}{\frac{{{X}-\mu}}{{\sigma}}}{<}{\frac{{{205.1}-{198.8}}}{{{69.2}}}}\right)}\)
\(\displaystyle={P}{\left(-{0.214}{<}{z}{<}{0.091}\right)}\)
\(\displaystyle={P}{\left({z}{<}{0.091}\right)}-{P}{\left({z}{<}-{0.214}\right)}\)
The probability of z less than 0.091 can be obtained using the excel formula “=NORM.S.DIST(0.091,TRUE)”. The probability value is 0.5363.
The probability of z less than –0.214 can be obtained using the excel formula “=NORM.S.DIST(–0.214,TRUE)”. The probability value is 0.4153.
The required probability value is,
\(\displaystyle{P}{\left({184}{<}{X}{<}{205.1}\right)}={P}{\left({z}{<}{0.091}\right)}-{P}{\left({z}{<}-{0.214}\right)}\)
\(\displaystyle={0.5363}−{0.4153}={0.1210}\)
Thus, the probability that a single randomly selected value is between 184 and 205.1 is 0.1210.

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