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The Globe Fishery packs shrimp that weigh more than 1.94 ounces each in packages marked "large" and shrimp that weigh less than 0.98 ounces each into packages marked "small"; the remainder are packed in "medium" size packages. If a day's catch showed that 19.77 percent of the shrimp were large and 6.06 percent were small, determine the mean and the standard deviation for the shrimp weights (in oz). Assume that the shrimps' weights are normally distributed. (Round your answers to one decimal place.)

Mean: ______ oz

Standard Deviation: ______ oz

Answer :

Final answer:

Calculate the mean and standard deviation of the shrimp weights based on given percentages. Use the provided formula and distribution information to determine the mean and standard deviation.

Explanation:

Mean = 1.46 oz, Standard Deviation = 0.48 oz

  1. Calculate the mean weight: (0.19 * 1.94) + (0.06 * 0.98) + (0.75 * X) = 1.46
  2. Calculate the variance of the weights using the formula: Variance = (0.19 * (1.94 - 1.46)^2) + (0.06 * (0.98 - 1.46)^2) + (0.75 * (X - 1.46)^2)
  3. Finally, calculate the standard deviation by taking the square root of the variance calculated in the previous step.

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