Question

A population of values has a normal distribution with \mu=77 and \sigma=32.2.

Random variables
ANSWERED
asked 2020-12-05

A population of values has a normal distribution with \(\displaystyle\mu={77}\) and \(\displaystyle\sigma={32.2}\). You intend to draw a random sample of size \(\displaystyle{n}={15}\)
Find the probability that a single randomly selected value is between 59.5 and 98.6. \(\displaystyle{P}{\left({59.5}{<}{X}{<}{98.6}\right)}=\)?
Write your answers as numbers accurate to 4 decimal places.

Expert Answers (1)

2020-12-06

From the given information, \(\displaystyle\mu={77}\), \(\displaystyle\sigma={32.2}\) and the sample size=15.
Consider,
\(\displaystyle{P}{\left({59.5}{<}{X}{<}{98.6}\right)}={P}{\left({\frac{{{59.5}-\mu}}{{\sigma}}}{<}{\frac{{{X}-\mu}}{{\sigma}}}{<}{\frac{{{98.6}-\mu}}{{\sigma}}}\right)}\)
\(\displaystyle={P}{\left({\frac{{{59.5}-{77}}}{{{32.2}}}}{<}{z}{<}{\frac{{{98.6}-{77}}}{{{32.2}}}}\right)}\)
\(\displaystyle={P}{\left(-{0.54348}{<}{z}{<}{0.6770807}\right)}\)
\(\displaystyle={P}{\left({z}{<}{0.6770807}\right)}-{P}{\left({z}{<}-{0.54348}\right)}\)
\(=0.748828-0.2934=0.4554 \begin{bmatrix}From\ the\ Excel\ function \\=NORM.DIST(0.6770807,0,1,TRUE) \\=NORM.DIST(-0.54348,0,1,TRUE) \end{bmatrix}\)
Thus, the probability that a single randomly selected value is between 59.5 and 98.6 is 0.4554.

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