For a normally distributed population, approximately 68% of the population lies

Question

For a normally distributed population, approximately 68% of the population lies within 1 standard deviation of the mean and approximately 95% of the population lies within 2 standard deviations of the mean. In a certain environmental study, the weights of 10,000 fish are normally distributed with a mean of 12.5 ounces and a standard deviation of 4.2 ounces. Approximately what percent of the 10,000 fish have weights between 16.7 and 20.9 ounces? Please have your final answer be a percent. Explain your answer.

Explanation / Answer

THIS WILL BE HELPFUL FOR YOU AND PL RATE ME FIRST Total number of fishes between (12. 5 - 8.4) and (12. 5 + 8.4) = 95% Total number of fishes between 12.5 and (12. 5 + 8.4) = 47.5% So, there are 0.475*10000=4750 fishes between 12.5 and 20.9. Total number of fishes between (12. 5 - 4.2) and (12. 5 + 4.2) = 68% Total number of fishes between 12.5 and (12. 5 + 4.2) = 34% So, there are 0.34*10000=3400 fishes between 12.5 and 16.7. Then the number of fishes between 16.4 and 20.9=4750-3400=1350

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