A population of values has a normal distribution with μ=200 and σ=31.9. You intend to...

The probability that a single randomly selected value is less than 226.9 is,
$P\left(X<226.9\right)=P\left(\frac{X-\mu }{\sigma }<\frac{226.9-200}{31.9}\right)$
$=P\left(z<\frac{26.9}{31.9}\right)$
$=P\left(z<0.843\right)$
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,
$P\left(X<226.9\right)=P\left(z<0.843\right)=0.8004$
Thus, the probability that a single randomly selected value is less than 226.9 is 0.8004.

Answer is given below (on video)