\text{1. Convert the raw scores to the standard normal variable Z:} Z = \frac{X - \mu}{\sigma}
\text{For X = 95:} Z = \frac{95 - 100}{15} = -\frac{1}{3}
\text{For X = 110:} Z = \frac{110 - 100}{15} = \frac{2}{3}
\text{2. Find the probability for these Z values using the standard normal distribution table:}
P(Z
P(Z
\text{3. Compute the difference to find the percentage of the population within this score range:}
P(-\frac{1}{3}
P(-\frac{1}{3}
\text{Therefore, the percentage of the population that would obtain a coefficient between 95 and 110 is 37.81\%}