Study SimplifyStudy Simplify

Class 12 Statistics Notes · GSEB

Inverse Problems

Inverse Problems — work backwards from a given probability to find x, z, or the parameters μ and σ, with a gated step-by-step solver. GSEB Class 12 Commerce Statistics notes covering cut-offs, percentiles, and simultaneous equations.

Last updated: 23 Sep 2026

Notes

Forward vs Inverse Problems

Inverse Problem
A normal-distribution sum in which the probability (or percentage) is GIVEN and you must work backwards to find the value of X — a cut-off mark, an income limit, a decile, or even the parameters μ and σ themselves.

Forward (what you already know)

Marks: X ~ N(60, 8²). Find P(X ≥ 70).

X → z = (70−60)/8 = 1.25 → table → 0.5 − 0.3944 = 0.1056

Answer: a probability.

Inverse (this topic)

Marks: X ~ N(60, 8²). Top 10% get a scholarship — find the cut-off.

0.10 → z = 1.28 → x = 60 + 1.28(8) = 70.24

Answer: a value of X.

Forward chain

X  ⟶  Z  ⟶  areaX \;\longrightarrow\; Z \;\longrightarrow\; \text{area}

Inverse chain

area  ⟶  Z  ⟶  X=μ+Zσ\text{area} \;\longrightarrow\; Z \;\longrightarrow\; X = \mu + Z\sigma

Why the table reads backwards

The GSEB Z-table prints the area between 0 and Z. Going forward you know Z and read the area; going inverse you know the area and must hunt for it inside the table — then fix the sign from the diagram. Every inverse sum in this chapter is that one skill, wrapped in a different story.

Step 1 — Find z From the Given Area

Choose how the area is described, enter it, and watch the book-style nearest-before / nearest-after / average search pick the Z-score.

Reverse Z-Lookup — area given, find z
0 to Z = 0.3925−3−2−10123+1.24−1.24x

Step 1 — reduce to a 0-to-Z area

P(0 ≤ Z ≤ z) = 0.3925

Already a 0-to-Z area — the table reads it directly.

Step 2 — book-style table search

From tableAreaZ-score
Exact table entry0.39251.24

Step 3 — final Z-score

z = ±1.24

Two answers — one right of 0, one left. The diagram in your sum decides which to use.

The interpolation rule (exactly how the textbook picks)

When the given area is not printed in the table, compare three candidates — nearest before, nearest after, and their average — and take whichever is closest to your area. That is how the book reaches 1.645 for 0.45, 1.28 for 0.40, and 0.675 for 0.25.

Step 2 — Convert z Back to x

The master inverse formula

x=μ+zσx = \mu + z\sigma

P(X ≤ x) = p (bottom p%)

Find z for |p − 0.5| area; if p < 0.5 the z is negative. Then x = μ + zσ — a cut-off below the mean.

P(X ≥ x) = p (top p%)

If p < 0.5, z is positive: the cut-off sits above μ — like a scholarship or merit-list threshold.

P(μ ≤ X ≤ x) = p

The area starts at the mean — so p is the 0-to-Z area. Direct table lookup, z positive. Its mirror P(x ≤ X ≤ μ) = p gives the same area with z negative — use the solver's matching button.

Middle p% → x₁ and x₂

Split: each half = p/2. Find z once, then x₁ = μ − zσ and x₂ = μ + zσ — symmetric about the mean.

The #1 inverse mistake: the sign of z

"Bottom 10% of salaries" does not mean z = +1.28. Areas below 0.5 always land left of the mean → z = −1.28 → x = μ − 1.28σ. If your cut-off came out on the wrong side of μ, you flipped the sign. Sketch the shade before you compute.

Three checkpoints, exactly like the textbook: reduce to a 0-to-Z area → back-lookup with the interpolation table → x = μ + zσ. Each gate unlocks the next.

Inverse-X Solver — probability in, x out
Book preset: μ = 60, σ = 8, P(X ≥ x) = 0.10 (scholarship cutoff, Illustration 11)
1

Reduce to a 0-to-Z area: what is P(0 ≤ Z ≤ z) for this problem?

2

Search the full 0-to-Z table below for your area — what is z?

3

Convert back: x = μ + zσ — what is the value of x?

μμ ± σμ ± σμ ± 2σμ ± 2σμ ± 3σμ ± 3σx

Percentiles and Deciles Are Inverses Too

Decile / Percentile
The value below which a given fraction of the observations lie. k-th decile: P(X ≤ D_k) = k/10 — D₃ cuts off the lowest 30%. n-th percentile: P(X ≤ P_n) = n/100 — P₆₀ cuts off the lowest 60%. Either way it is a left-tail inverse problem.

Decile

P(X≤Dk)=k10P(X \leq D_k) = \frac{k}{10}

Percentile

P(X≤Pn)=n100P(X \leq P_n) = \frac{n}{100}

Worked in one line of logic (Illustration 14)

Heights N(165, 10²): third decile → P(X ≤ D₃) = 3/10 = 0.30 → required area = 0.5 − 0.30 = 0.20 (left tail, so z is negative) → table gives |z| ≈ 0.525 → D₃ = 165 − 0.525(10) = 159.75 cm. Interpretation: 30% of the group is at or below 159.75 cm.

Step 3 — Finding μ and σ Themselves

Sometimes the probability condition is given but a parameter is missing. Rearranged Z once gives one unknown; two conditions give two equations.

Find Standard Deviation

σ=x−μz\sigma = \frac{x - \mu}{z}

Find Mean

μ=x−zσ\mu = x - z\sigma

Find Both Parameters

σ=x1−x2z1−z2,μ=x1−z1σ\sigma = \frac{x_1 - x_2}{z_1 - z_2}, \quad \mu = x_1 - z_1\sigma

One condition → one unknown

Bulb life: μ = 2040 h known, P(X ≥ 2150) = 0.0336. Table → P(0≤Z≤z) = 0.4664 → z = 1.83. Then 1.83 = (2150 − 2040)/σ → σ = 60.11 h (σ² = 3613.21).

Two conditions → solve together

City temperatures: P(X ≥ 31) = 0.3085 → z₁ = 0.5; P(X ≤ 27) = 0.0668 → z₂ = −1.5. Two equations 31 − μ = 0.5σ and 27 − μ = −1.5σ → subtract → 4 = 2σ → σ = 2, μ = 30°C.

Real-life parallel: two percentiles pin a distribution

F-6 in the exercise does the same thing with shop bills — 7.78% below ₹3,590 and 94.52% below ₹5,100 give two z-values, two equations, and both parameters. Auditors and pricing teams actually use this "two-point calibration" to fit normal models to real data.

Solved Examples

Solved Example

Problem

Marks of 200 students ~ N(60, 8²). The top 10% get a scholarship. Find the minimum marks for the scholarship.

Solution

Minimum marks ≈ 70.24 (about 21 of the 200 students qualify)

Solved Example

Problem

Monthly income ~ N(₹15,000, ₹4000²). Find the income range of the middle 60% of employees.

Solution

Middle 60% earn between ₹11,640 and ₹18,360

Solved Example

Problem

If P(0 ≤ Z ≤ z) = 0.3925, find the possible values of z.

Solution

z = +1.24 or z = −1.24 (the diagram decides which)

Solved Example

Problem

Salary ~ N(μ, 1500²). P(X ≤ 1000) = 0.0918. Find μ.

Solution

μ = ₹1,199.5

Key Takeaways

Key Takeaways

  • Inverse problems flip the chain: given the area, back-lookup z in the table, then x = μ + zσ — every time.
  • Always reduce the story to a 0-to-Z area first using 0.5 and symmetry; that single skill covers left tails, right tails, and mean-based ranges.
  • When the area is between two table entries, compare nearest-before, nearest-after and their average — closest wins (this is exactly how the book gets 1.645, 1.28 and 0.675).
  • The sign of z is decided by the diagram, never by the size of the probability alone — bottom p% and top p% of equal size sit on opposite sides of μ.
  • One probability condition gives one parameter equation; two conditions (or two X-Z pairs) let you solve for both μ and σ simultaneously — the hardest inverse sums in Exercise 3.
  • Deciles and percentiles are left-tail inverses — P(X ≤ D_k) = k/10 for deciles, P(X ≤ P_n) = n/100 for percentiles — then apply the same three steps.

Practice

  1. Salary ~ N(₹10,000, ₹2000²). Find (a) the maximum salary of the lowest 20% of workers, and (b) the minimum salary of the top 10%.
  2. P(Z ≤ z) = 0.15 and P(Z ≥ z) = 0.75 — find z in both cases and explain why the answers match.
  3. Heights ~ N(165, 10²). Find the 60th percentile and interpret it in one sentence.
  4. Bulbs: P(X ≥ 2150) = 0.0336 with μ = 2040. Find σ (Illus 15 — try it before peeking at the solution).

Explore Further