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Inverse Normal Calculations. Consider a population of crabs where the length of a shell, X mm, is normally distributed with a mean of 70mm and a standard.

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Presentation on theme: "Inverse Normal Calculations. Consider a population of crabs where the length of a shell, X mm, is normally distributed with a mean of 70mm and a standard."— Presentation transcript:

1 Inverse Normal Calculations

2 Consider a population of crabs where the length of a shell, X mm, is normally distributed with a mean of 70mm and a standard deviation of 10mm. A biologist wants to protect the population by allowing only the largest 5% of crabs to be harvested. He therefore asks the question: “95% of the crabs have lengths less than what?” The biologist needs to find k such that P(X < k) = 0.95. The number k is known as a quantile, and in this case, the 95% quantile.

3 Inverse Normal Calculations

4 Example 1. Consider a population of crabs where the length of a shell, X mm, is normally distributed with a mean of 70mm and a standard deviation of 10mm. A biologist wants to protect the population by allowing only the largest 5% of crabs to be harvested. He therefore asks the question: “95% of the crabs have lengths less than what?” invNorm(0.95, 70, 10) = 86.4485 = 86.4mm

5 Inverse Normal Calculations Example 2. The volume of cartons of milk is normally distributed with a mean of 995 ml and a standard deviation of 5 ml. It is known that 10% of cartons have a volume less than x ml. Find the value of x. invNorm(0.1, 995, 5) = 988.5922 = 989 ml

6 Inverse Normal Calculations To perform inverse normal calculations on the calculator, we must enter the area to the left of k. If P(X > k) = p, then P(X < k) = 1 - p

7 Inverse Normal Calculations Example 3. A university professor determines that 80% of this year’s history students should pass the final exam. The exam results were approximately normally distributed with a mean of 62 and a standard deviation of 12. Find the lowest score necessary to pass the exam. P(X > k) = 0.8, which is the same thing as saying P(X < k) = 1- 0.8 = 0.2 invNorm(0.2, 62, 12) = 51.9005 = 51.9

8 Inverse Normal Calculations Example 4. The weights of pears are normally distributed with a mean of 110g and a standard deviation of 8g. a.Find the percentage of pears that weighs between 100g and 130g. normalcdf(100, 130, 110, 8) = 0.88814 = 88.8%

9 Inverse Normal Calculations Example 4. The weights of pears are normally distributed with a mean of 110g and a standard deviation of 8g. b. It is known that 8% of the pears weigh more than m g. Find the value of m. invNorm(0.92, 110, 8) = 121.2405 = 121 g

10 Inverse Normal Calculations Example 4. The weights of pears are normally distributed with a mean of 110g and a standard deviation of 8g. c. 250 pears are weighed. Calculate the expected number of pears that weigh less than 105g. normalcdf(-1E99, 105, 110, 8) = 0.265985 = 0.266 0.266 x 250 = 66 or 67 pears (depending on rounding)


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