Inverse Normal Calculations

Inverse Normal CalculationsConsider 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.

Inverse Normal Calculations

When finding quantiles, we are given a probability and are asked to calculate the corresponding measurement. This is the inverse of finding probabilities, and we use the inverse normal function on our calculator.

What to type:“2nd” “VARS”“invNorm(”Enter area to the left, μ

Inverse Normal CalculationsExample 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

Inverse Normal CalculationsExample 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

Inverse Normal CalculationsTo 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

Inverse Normal CalculationsExample 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 = 0.2invNorm(0.2, 62, 12)= 51.9005 = 51.9

Inverse Normal CalculationsExample 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%

Inverse Normal CalculationsExample 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

Inverse Normal CalculationsExample 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 = 0.2660.266 x 250 = 66 or 67 pears(depending on rounding)