Answer :
The mean shrimp weight is approximately 1.1 ounces, and the standard deviation is approximately 0.4110 ounces (rounded to one decimal place).
1. The percentage of large shrimp is 19.77%, which means the remaining percentage of shrimp are medium-sized. Since the total percentage must add up to 100%, the percentage of medium-sized shrimp is:
100% - 19.77% - 6.06% = 74.17%
2. Now, we'll convert these percentages into decimal probabilities:
The probability of a shrimp being large (L) = 0.1977
The probability of a shrimp being medium (M) = 0.7417
The probability of a shrimp being small (S) = 0.0606
3. Next, we'll use these probabilities along with the shrimp weights to calculate the mean (expected value) and standard deviation:
Let X be the random variable representing the shrimp weights (in ounces).
The mean (μ) can be calculated as:
μ = (Probability of L * Weight of L) + (Probability of M * Weight of M) + (Probability of S * Weight of S)
μ = (0.1977 * 1.6 oz) + (0.7417 * 1.0 oz) + (0.0606 * 0.6 oz)
The standard deviation (σ) can be calculated as:
σ = sqrt((Probability of L * (Weight of L - Mean)^2) + (Probability of M * (Weight of M - Mean)^2) + (Probability of S * (Weight of S - Mean)^2))
σ = sqrt((0.1977 * (1.6 - μ)^2) + (0.7417 * (1.0 - μ)^2) + (0.0606 * (0.6 - μ)^2))
4. Now, we can calculate the mean and standard deviation:
First, calculate the mean (μ):
μ = (0.1977 * 1.6 oz) + (0.7417 * 1.0 oz) + (0.0606 * 0.6 oz)
μ = 0.31632 oz + 0.7417 oz + 0.03636 oz
μ = 1.09438 oz (rounded to one decimal place)
Now, calculate the standard deviation (σ):
σ = sqrt((0.1977 * [tex](1.6 - 1.09438)^2[/tex]) + (0.7417 *[tex](1.0 - 1.09438)^2[/tex]) + (0.0606 * [tex](0.6 - 1.09438)^2)[/tex])
σ = sqrt(0.113837 + 0.006982 + 0.047947)
σ = sqrt(0.168766)
σ ≈ 0.4110 oz (rounded to four decimal places)
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Complete Question is:
You may need to use the appropriate appendix table or technology to anwer this gutstion. are packed in "medum" size packages. If a day's catch showed that 19.77 percent of the shrimp were large and 6.06 penoent were imall, detemine the mean and the ifanded efvithe for the. shrimp welghts (in oq). Astume that the shrimps' weights are normally distributed. (Round your answers to one dechal place.) Find Mean and standard deviation.
Final answer:
The mean and standard deviation for the shrimp weights can be calculated by multiplying the percentage of each size category by the corresponding weight and summing the weighted values. For a day's catch with 19.77% large shrimp, 74.17% medium shrimp, and 6.06% small shrimp, the mean and standard deviation can be determined using this method.
Explanation:
To calculate the mean and standard deviation for the shrimp weights, we need to use the given information. We know that 19.77% of the shrimp were large and 6.06% were small. This means that the remaining percentage, 100% - 19.77% - 6.06% = 74.17%, represents the medium-sized shrimp.
Since the weights of the shrimp are normally distributed, we can assume a bell-shaped curve. To find the mean, we multiply the percentage of each size category by the corresponding weight:
- Large shrimp: 19.77% * weight of large shrimp
- Medium shrimp: 74.17% * weight of medium shrimp
- Small shrimp: 6.06% * weight of small shrimp
Similarly, to find the standard deviation, we need to calculate the weighted standard deviation for each size category.
Once we have the mean and standard deviation for each size category, we can calculate the overall mean and standard deviation by summing the weighted values.
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