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Background The growth rate of abalone can be monitored using a statistic known ...
May 23, 2024
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Solution by Steps
step 1
We start by calculating the probabilities for the normal distribution with mean values given in the table and standard deviation σ=0.215\sigma = 0.215
step 2
For each mean SGR value, we calculate the cumulative distribution function (CDF) at 0.560.56 to determine the probability of choosing Option 1
step 3
Using the CDF values, we can determine the probabilities for Options 2 and 3
step 4
For the mean SGR of 0.60.6, the CDF at 0.560.56 is 0.41490.4149. Therefore, the probability of choosing Option 1 is 0.41490.4149
step 5
The probability of choosing Option 2 is the difference between the CDF values at 0.600.60 and 0.560.56
step 6
The probability of choosing Option 3 is 1−CDF at 0.601 - \text{CDF at } 0.60
step 7
Repeat steps 4-6 for each mean SGR value in the table
step 8
For the proportion of saleable abalone, we calculate 1−CDF at 0.561 - \text{CDF at } 0.56 for each mean SGR value
step 9
For the binomial distribution, we use the proportion of saleable abalone to calculate the probabilities of having at least 50,60,70,80,50, 60, 70, 80, and 9090 acceptable abalone out of 100100
step 10
For the profit probability, we use the binomial distribution results to determine the probability of making a profit under Options 2 and 3
Answer
The probabilities and proportions for each mean SGR value are calculated using the normal and binomial distributions.
Key Concept
Normal and Binomial Distributions
Explanation
The normal distribution is used to model the SGR values, and the binomial distribution is used to calculate the probabilities of having a certain number of acceptable abalone.
Generated Graph
Solution by Steps
step 1
To determine the probabilities for each option based on the mean SGR for the tank, we need to use the normal distribution with the given mean and standard deviation
step 2
For each mean SGR value, we calculate the probability that the SGR is less than 0.56 (Option 1), between 0.56 and 0.60 (Option 2), and greater than 0.60 (Option 3)
step 3
Using the normal distribution with mean μ=0.6 \mu = 0.6 and standard deviation σ=0.215 \sigma = 0.215 , we calculate P(X < 0.56) for Option 1: P(X < 0.56) = 0.426204
step 4
For Option 2, we calculate P(0.56≤X≤0.60) P(0.56 \leq X \leq 0.60) : P(0.56 \leq X \leq 0.60) = P(X \leq 0.60) - P(X < 0.56) = 0.5 - 0.426204 = 0.073796
step 5
For Option 3, we calculate P(X > 0.60) : P(X > 0.60) = 1 - P(X \leq 0.60) = 1 - 0.5 = 0.5
step 6
Repeat the above steps for each mean SGR value (0.64, 0.68, 0.72, 0.76, 0.8, 0.84) using the normal distribution with the respective mean and standard deviation σ=0.215 \sigma = 0.215
step 7
For mean SGR μ=0.64 \mu = 0.64 : P(X < 0.56) = 0.354912 P(0.56≤X≤0.60)=0.5−0.354912=0.145088 P(0.56 \leq X \leq 0.60) = 0.5 - 0.354912 = 0.145088 P(X > 0.60) = 1 - 0.5 = 0.5
step 8
For mean SGR μ=0.68 \mu = 0.68 : P(X < 0.56) = 0.288375 P(0.56≤X≤0.60)=0.5−0.288375=0.211625 P(0.56 \leq X \leq 0.60) = 0.5 - 0.288375 = 0.211625 P(X > 0.60) = 1 - 0.5 = 0.5
step 9
For mean SGR μ=0.72 \mu = 0.72 : P(X < 0.56) = 0.228382 P(0.56≤X≤0.60)=0.5−0.228382=0.271618 P(0.56 \leq X \leq 0.60) = 0.5 - 0.228382 = 0.271618 P(X > 0.60) = 1 - 0.5 = 0.5
step 10
For mean SGR μ=0.76 \mu = 0.76 : P(X < 0.56) = 0.176125 P(0.56≤X≤0.60)=0.5−0.176125=0.323875 P(0.56 \leq X \leq 0.60) = 0.5 - 0.176125 = 0.323875 P(X > 0.60) = 1 - 0.5 = 0.5
step 11
For mean SGR μ=0.8 \mu = 0.8 : P(X < 0.56) = 0.132151 P(0.56≤X≤0.60)=0.5−0.132151=0.367849 P(0.56 \leq X \leq 0.60) = 0.5 - 0.132151 = 0.367849 P(X > 0.60) = 1 - 0.5 = 0.5
step 12
For mean SGR μ=0.84 \mu = 0.84 : P(X < 0.56) = 0.0964026 P(0.56≤X≤0.60)=0.5−0.0964026=0.403597 P(0.56 \leq X \leq 0.60) = 0.5 - 0.0964026 = 0.403597 P(X > 0.60) = 1 - 0.5 = 0.5
Answer
The probabilities for each option based on the mean SGR for the tank are as follows: - For mean SGR 0.6: Option 1: 0.426204, Option 2: 0.073796, Option 3: 0.5 - For mean SGR 0.64: Option 1: 0.354912, Option 2: 0.145088, Option 3: 0.5 - For mean SGR 0.68: Option 1: 0.288375, Option 2: 0.211625, Option 3: 0.5 - For mean SGR 0.72: Option 1: 0.228382, Option 2: 0.271618, Option 3: 0.5 - For mean SGR 0.76: Option 1: 0.176125, Option 2: 0.323875, Option 3: 0.5 - For mean SGR 0.8: Option 1: 0.132151, Option 2: 0.367849, Option 3: 0.5 - For mean SGR 0.84: Option 1: 0.0964026, Option 2: 0.403597, Option 3: 0.5
Key Concept
Normal Distribution Probabilities
Explanation
The probabilities for each option are calculated using the normal distribution with the given mean and standard deviation. The cumulative distribution function (CDF) is used to find the probabilities for the specified ranges.
background is same as the last question
Solution by Steps
step 1
Identify the mean SGR values from the table: 0.6, 0.64, 0.68, 0.72, 0.76, 0.8, 0.84
step 2
The minimum acceptable SGR level is 0.56. We need to compute the proportion of abalone in the tank with an SGR level above 0.56 for each mean SGR value
step 3
For the mean SGR of 0.6, the proportion saleable is given as 0.574
step 4
For the other mean SGR values, we need to calculate the proportion saleable. Assuming a normal distribution, we can use the Z-score formula: Z=X−μσ Z = \frac{X - \mu}{\sigma} where X X is the SGR value, μ \mu is the mean SGR, and σ \sigma is the standard deviation
step 5
For the mean SGR of 0.64, calculate the Z-score for 0.56: Z=0.56−0.64σ Z = \frac{0.56 - 0.64}{\sigma} Assuming a standard deviation (σ) of 0.1 (example value), the Z-score is: Z=0.56−0.640.1=−0.8 Z = \frac{0.56 - 0.64}{0.1} = -0.8 Using the Z-table, the proportion saleable is approximately 0.7881
step 6
Repeat the calculation for other mean SGR values (0.68, 0.72, 0.76, 0.8, 0.84) using the same method
step 7
For batches of 100 abalone, compute the probabilities that the number of acceptable abalone is at least 50, 60, 70, 80, and 90. Use the binomial distribution formula: P(X≥k)=∑i=kn(ni)pi(1−p)n−i P(X \geq k) = \sum_{i=k}^{n} \binom{n}{i} p^i (1-p)^{n-i} where n n is the total number of abalone, k k is the number of acceptable abalone, and p p is the proportion saleable
step 8
For the mean SGR of 0.6 and at least 50 acceptable abalone: P(X≥50)=∑i=50100(100i)(0.574)i(1−0.574)100−i P(X \geq 50) = \sum_{i=50}^{100} \binom{100}{i} (0.574)^i (1-0.574)^{100-i} Using a binomial calculator, the probability is approximately 0.944
step 9
Repeat the calculation for other mean SGR values and different numbers of acceptable abalone (60, 70, 80, 90)
step 10
Compute the probabilities needed to complete the table showing the probability of making a profit under Option 2 and Option 3. For Option 2, at least 90% of the abalone must have SGR levels above 0.56. For Option 3, at least 70% of the abalone must have SGR levels above 0.56
step 11
For Option 2 and a mean SGR of 0.6: P(X≥90)=∑i=90100(100i)(0.574)i(1−0.574)100−i P(X \geq 90) = \sum_{i=90}^{100} \binom{100}{i} (0.574)^i (1-0.574)^{100-i} Using a binomial calculator, the probability is approximately 0
step 12
For Option 3 and a mean SGR of 0.8: P(X≥70)=∑i=70100(100i)(0.8)i(1−0.8)100−i P(X \geq 70) = \sum_{i=70}^{100} \binom{100}{i} (0.8)^i (1-0.8)^{100-i} Using a binomial calculator, the probability is approximately 0.998
Answer
The probabilities for making a profit under Option 2 and Option 3 are calculated based on the mean SGR values and the required proportions of acceptable abalone. For Option 3 with a mean SGR of 0.8, the probability of making a profit is 0.998.
Key Concept
Binomial Distribution
Explanation
The binomial distribution is used to calculate the probability of a certain number of successes in a fixed number of trials, given the probability of success in each trial. In this case, it helps determine the likelihood of having a certain number of acceptable abalone in a batch.
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