Given that the lengths of male snakes are normally distributed, we can use the standard normal distribution to find the mean \(\mu\) and standard deviation \(\sigma\).
From the problem, 32 out of 200 snakes have lengths less than 45 cm. This corresponds to a probability of \(\frac{32}{200} = 0.16\). Using the standard normal distribution table, this probability corresponds to a z-score of approximately \(-0.994\).
Similarly, 17 out of 200 snakes have lengths more than 56 cm, corresponding to a probability of \(\frac{17}{200} = 0.085\). This corresponds to a z-score of approximately \(1.372\).
We have the following standardization equations:
\(\frac{45 - \mu}{\sigma} = -0.994\)
\(\frac{56 - \mu}{\sigma} = 1.372\)
Solving these two equations simultaneously, we first eliminate \(\mu\) by subtracting the first equation from the second:
\(\frac{56 - \mu}{\sigma} - \frac{45 - \mu}{\sigma} = 1.372 + 0.994\)
\(\frac{11}{\sigma} = 2.366\)
\(\sigma = \frac{11}{2.366} \approx 4.65\)
Substituting \(\sigma = 4.65\) back into one of the original equations to find \(\mu\):
\(\frac{45 - \mu}{4.65} = -0.994\)
\(45 - \mu = -0.994 \times 4.65\)
\(\mu = 45 + 4.6251 \approx 49.6\)
Thus, the estimates for the mean and standard deviation are \(\mu = 49.6\) and \(\sigma = 4.65\).
