1. The problem asks to find the percentage of data between the z-scores $-0.14$ and $0.09$ in a standard normal distribution. 2. We use the standard normal distribution table (Z-table) or a calculator to find the cumulative probabilities for each z-score. 3. The cumulative probability for $z = -0.14$ is approximately $P(Z \leq -0.14) = 0.4443$. 4. The cumulative probability for $z = 0.09$ is approximately $P(Z \leq 0.09) = 0.5359$. 5. To find the percentage of data between these z-scores, subtract the smaller cumulative probability from the larger one: $$ P(-0.14 \leq Z \leq 0.09) = P(Z \leq 0.09) - P(Z \leq -0.14) = 0.5359 - 0.4443 = 0.0916 $$ 6. Convert this to a percentage by multiplying by 100: $$ 0.0916 \times 100 = 9.16\% $$ 7. Therefore, approximately 9.16% of the data lies between the z-scores $-0.14$ and $0.09$.