Question # A population of values has a normal distribution with \mu=26.8 and \sigma=33.8. You intend to draw a random sample of size n=89.

Random variables
ANSWERED A population of values has a normal distribution with $$\displaystyle\mu={26.8}$$ and $$\displaystyle\sigma={33.8}$$. You intend to draw a random sample of size $$\displaystyle{n}={89}$$.
Find the probability that a single randomly selected value is between 17.1 and 25.
$$\displaystyle{P}{\left({17.1}{<}{X}{<}{25}\right)}=$$? 2020-10-21

Step 1
It is given that a population of values distributed normally with mean 26.8 and standard deviation 33.8. The sample size is 89.
Step 2
Calculate the probability that a single randomly selected value is between 17.1 and 25 is as follows:
$$\displaystyle{P}{\left({17.1}{<}{X}{<}{25}\right)}={P}{\left({\frac{{{17.1}-\mu}}{{\sigma}}}{<}{\frac{{{X}-\mu}}{{\sigma}}}{<}{\frac{{{25}-\mu}}{{\sigma}}}\right)}$$
$$\displaystyle={P}{\left({\frac{{{17.1}-{26.8}}}{{{33.8}}}}{<}{Z}{<}{\frac{{{25}-{26.8}}}{{{33.8}}}}\right)}$$
$$\displaystyle={P}{\left(-{0.287}{<}{Z}{<}-{0.053}\right)}$$
$$\displaystyle={P}{\left({Z}{<}-{0.053}\right)}-{P}{\left({Z}{<}-{0.287}\right)}{\left({b}{e}{g}\in{\left\lbrace{m}{a}{t}{r}{i}{x}\right\rbrace}{U}{\sin{{g}}}\ {E}{x}{c}{e}{l}\ {f}{u}{n}{c}{t}{i}{o}{n}{s},\backslash{P}{\left({Z}{<}-{0.053}\right)}={0.4789}\backslash{P}{\left({Z}{<}-{0.287}\right)}={0.3871}{e}{n}{d}{\left\lbrace{m}{a}{t}{r}{i}{x}\right\rbrace}\right)}$$
$$\displaystyle={0.4789}-{0.3871}={0.0918}$$
Therefore, the value of $$\displaystyle{P}{\left({17.1}{<}{X}{<}{25}\right)}={0.0918}$$.