Given that 69% of the distribution is less than 28, we find the corresponding z-score from the standard normal distribution table, which is approximately 0.496. Thus, we have:
\(28 - \mu = 0.496\sigma\)
Similarly, for 90% less than 35, the z-score is approximately 1.282. Thus, we have:
\(35 - \mu = 1.282\sigma\)
We now have two equations:
- \(28 - \mu = 0.496\sigma\)
- \(35 - \mu = 1.282\sigma\)
Subtract the first equation from the second:
\((35 - \mu) - (28 - \mu) = 1.282\sigma - 0.496\sigma\)
\(7 = 0.786\sigma\)
Solving for \(\sigma\):
\(\sigma = \frac{7}{0.786} \approx 8.91\)
Substitute \(\sigma = 8.91\) back into the first equation:
\(28 - \mu = 0.496 \times 8.91\)
\(28 - \mu = 4.42\)
\(\mu = 28 - 4.42 = 23.6\)
Thus, the mean \(\mu = 23.6\) and the standard deviation \(\sigma = 8.91\).
