Given that the weights are normally distributed, we can use the standard normal distribution to find the mean \(\mu\) and standard deviation \(\sigma\).
For 20% of pandas weighing more than 121 kg, the corresponding z-score is \(z = 0.842\). Using the standardization formula:
\(z = \frac{121 - \mu}{\sigma}\)
Substituting the values, we get:
\(0.842 = \frac{121 - \mu}{\sigma}\)
So, \(0.842\sigma = 121 - \mu\).
For 71.9% of pandas weighing more than 102 kg, the corresponding z-score is \(z = -0.58\). Using the standardization formula:
\(z = \frac{102 - \mu}{\sigma}\)
Substituting the values, we get:
\(-0.58 = \frac{102 - \mu}{\sigma}\)
So, \(-0.58\sigma = 102 - \mu\).
We now have two equations:
- \(0.842\sigma = 121 - \mu\)
- \(-0.58\sigma = 102 - \mu\)
Solving these simultaneously, we eliminate \(\mu\) and solve for \(\sigma\):
\(0.842\sigma + \mu = 121\)
\(-0.58\sigma + \mu = 102\)
Subtract the second equation from the first:
\((0.842 + 0.58)\sigma = 121 - 102\)
\(1.422\sigma = 19\)
\(\sigma = \frac{19}{1.422} \approx 13.4\)
Substitute \(\sigma = 13.4\) back into one of the original equations to find \(\mu\):
\(0.842 \times 13.4 + \mu = 121\)
\(11.2908 + \mu = 121\)
\(\mu = 121 - 11.2908\)
\(\mu \approx 110\)
Thus, the mean \(\mu\) is 110 and the standard deviation \(\sigma\) is 13.4.
