91Ó°ÊÓ

We are given that each fish caught is equally likely to be any one of the 4 types. According to the problem, let's denote the events: \(A_i\): Catching the \(i\)th distinct fish type for the first time, where \(i = 1, 2, 3, 4\). The probability of each event happening is: \(P(A_1) = 1\) \(P(A_2) = \frac{3}{4}\) \(P(A_3) = \frac{2}{4}\) \(P(A_4) = \frac{1}{4}\)

Using the probabilities from Step 1, we have: \(E(Y) = 1 \cdot P(A_1) + 2 \cdot P(A_2) + 3 \cdot P(A_3) + 4 \cdot P(A_4)\) \(E(Y) = 1 \cdot 1 + 2 \cdot \frac{3}{4} + 3 \cdot \frac{2}{4} + 4 \cdot \frac{1}{4}\) \(E(Y) = 1 + \frac{3}{2} + 3 + 1 = 8.5\)

We need to find \(a\) and \(b\) such that \(P\{a \leq Y \leq b\} \geq 0.9\). To do this, we can use the empirical rule to approximate the confidence interval. The empirical rule states that in a normal distribution, approximately: - 68% of the data falls within one standard deviation of the mean. - 95% of the data falls within two standard deviations of the mean. - 99.7% of the data falls within three standard deviations of the mean. Since we want at least 90% certainty, we can use two standard deviations from the mean. Note that this approximation should only be used if large numbers are involved (which might not be ideal for this case) and no other exact method is available. Let's assume the standard deviation \(\sigma(Y) = 2\). Thus the interval is: \((a, b) = (E(Y) - 2\sigma(Y), E(Y) + 2\sigma(Y))\) \((a, b) = (8.5 - 4, 8.5 + 4)\) \((a, b) = (4.5, 12.5)\)

We need to be at least 90% certain of obtaining at least one of each type using the one-sided Chebyshev inequality, which is given by: \[ P(Y \geq E(Y) - k\sigma(Y)) \geq 1 - \frac{1}{k^2}\] We want at least 90% certainty, which means: \[ 1 - \frac{1}{k^2} \geq 0.9\] \[ \frac{1}{k^2} \leq 0.1\] \[ k^2 \geq 10\] \[ k \geq \sqrt{10}\] Now, calculate \(Y\): \[ Y \geq E(Y) - k\sigma(Y)\] \[ Y \geq 8.5 - \sqrt{10} \cdot 2\] Approximately, we get: \[ Y \geq 8.5 - 6.32\] \[ Y \geq 2.18\] So, we need to plan on catching at least 3 fish to be at least 90% certain of obtaining at least one of each type using the one-sided Chebyshev inequality. Note that the answer may differ from part (a) since the approximation in part (a) was based on the empirical rule while part (b) used the one-sided Chebyshev inequality.