Question

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

Random variables

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. $$\displaystyle{P}{\left({59.5}{<}{X}{<}{98.6}\right)}=$$?

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.