Inverse Normal

Given the area, find the cut-off value.

Given the area, find the value

A normal probability question gives a value and asks for an area: what share of the scores are above 650? An inverse normal question runs the other way. It gives the area and asks for the value: above what score are the top 10%?

The top 10% of the area lies to the right of some z. A table gives areas to the left, so turn the question round: that z has 100% − 10% = 90% of the area to its left.

z

The shaded 90% of the area lies to the left of z = 1.28. The unshaded 10% to its right is the top 10%.

Read the table backward

Search the body of the table for 0.9000 and read off its z. In the row for 1.2, the column for 0.08 gives 0.8997 and the column for 0.09 gives 0.9015. 0.9000 is just past 0.8997, so z = 1.28 to two decimal places. More exactly z = 1.2816, which is what a calculator’s inverse normal function gives.

The usual slip is to look up the tail, 0.1000, instead of the area to the left. That gives a z near −1.28, at the bottom of the curve instead of the top.

0.060.070.080.091.10.87700.87900.88100.88301.20.89620.89800.89970.90151.30.91310.91470.91620.9177

In the row for 1.2, the area 0.9000 falls between 0.8997, at z = 1.28, and 0.9015, at z = 1.29, and much nearer the first.

Undo the standardizing

The z-score is on the standard scale, and the answer is wanted on the original one. Standardizing subtracted the mean and divided by σ, so undo the two moves in reverse order: multiply by σ, then add the mean back. From z = (x − μ)/σ, x = μ + zσ.

Test scores follow N(500, 100²). The top 10% starts at x = 500 + 1.28 × 100 = 628, so a score of 628 or more is in the top 10%. With the more exact z, 500 + 1.2816 × 100 = 628.2. Check by standardizing again: (628 − 500)/100 = 1.28.

score

The test scores, N(500, 100²). The shaded top 10% starts at 628, which is 1.28 standard deviations above the mean.

A lower tail, and a middle band

Below what score are the bottom 5%? The table gives 1.645 for 5% in the upper tail, and by symmetry the lower 5% lies below z = −1.645. That cut is below the mean, so its z is negative, and x = 500 + (−1.645) × 100 = 500 − 164.5 = 335.5.

Which scores make up the middle 95%? That leaves 2.5% in each tail, and 2.5% in a tail means z = 1.96. So the middle 95% runs from 500 − 1.96 × 100 = 304 to 500 + 1.96 × 100 = 696.

Finding the standard deviation

The same equation can be solved for σ or for μ. A bakery’s loaves have mean mass 800 g, and 5% of them weigh less than 780 g. The lowest 5% lies below z = −1.645, so 780 g must sit there: (780 − 800)/σ = −1.645.

Multiply both sides by σ: −20 = −1.645σ, so σ = 20/1.645 = 12.2 g.

The usual mistakes

Moving the wrong way from the mean. A positive z is above the mean and a negative z below it, so the bottom 5% is at 500 − 164.5, not 500 + 164.5.

Adding z itself to the mean. z counts standard deviations, so it is multiplied by σ first: 500 + 1.28 × 100, not 500 + 1.28.

Using 1.645 for a middle 95%. The 5% outside is split between two tails, 2.5% each, and that needs 1.96.

Worked example: A Machine Filling 500 g Bags of Rice: the Share That Is Underweight and the Mean It Must Be Set To

Question A machine fills bags of rice labeled 500 g. The mass of rice in a bag is normally distributed with standard deviation 4 g, and the mean can be set. (a) With the mean set at 505 g, find the proportion of bags that are underweight. (b) The law allows at most 2.5% of bags to be under 500 g. Find the lowest mean the machine can be set to.

  1. 1.Let M ∼ N(505, 42). A bag is underweight when M < 500, and z = 500 − 5054 = −1.25.

    gz500−1.255050N(505, 16)z = (500 − 505)/4 = −1.25
    gz500−1.255050N(505, 16)z = (500 − 505)/4 = −1.25
    With the mean at 505 g, the label mass 500 g sits at z = 500 − 5054 = −1.25.

  2. 2.(a) P(M < 500) = Φ(−1.25) = 1 − Φ(1.25) = 1 − 0.8944 = 0.1056, so about 10.6% of bags are underweight.

    gz500−1.255050N(505, 16)Φ(−1.25) = 1 − 0.8944 = 0.1056
    gz500−1.255050N(505, 16)Φ(−1.25) = 1 − 0.8944 = 0.1056
    (a) The shaded tail is P(M < 500) = 1 − Φ(1.25) = 0.1056: about 10.6% of bags are underweight.

  3. 3.Let the new mean be μ. The table gives Φ(1.96) = 0.975, so the lowest 2.5% of a normal distribution lies below z = −1.96. The mass 500 g must sit at that z: 500 − μ4 = −1.96.

    gz500−1.96mean0mean unknown, sd 4lowest 2.5%: z = −1.96(500 − mean)/4 = −1.96
    gz500−1.96mean0mean unknown, sd 4lowest 2.5%: z = −1.96(500 − mean)/4 = −1.96
    For only 2.5% below 500 g, the label must sit at z = −1.96, since Φ(1.96) = 0.975.

  4. 4.Multiply both sides by 4: 500 − μ = −7.84, so μ = 507.84.

    gz500−1.96507.840N(507.84, 16)multiply both sides by 4500 − mean = −7.84
    gz500−1.96507.840N(507.84, 16)multiply both sides by 4500 − mean = −7.84
    500 − μ4 = −1.96 gives 500 − μ = −7.84.

  5. 5.(b) The lowest mean is 507.84 g, about 507.8 g. Check: 500 − 507.844 = −1.96, and Φ(−1.96) = 1 − 0.975 = 0.025.

    gz500−1.96507.840N(507.84, 16)mean = 507.84 gcheck: (500 − 507.84)/4 = −1.96
    gz500−1.96507.840N(507.84, 16)mean = 507.84 gcheck: (500 − 507.84)/4 = −1.96
    (b) μ = 507.84 g. The curve has moved right until the tail below 500 g is 0.025.

Answer: (a) 0.1056; (b) 507.84 g

Common mistakes

  • Using z = +1.96, which gives μ = 500 − 7.84 = 492.16 g. A mean below the label would make most bags underweight; 500 g lies below the mean, so its z is negative.
  • Using z = −1.645, the value for 5% in one tail. The law allows 2.5% below 500 g, all in the lower tail, and that needs z = −1.96.

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