Normal distribution: standardization and tails

Objectives and prerequisites

Interpret normal-distribution parameters; standardize by the standard deviation; read CDF values and tails; calculate interval probabilities and quantiles; and combine independent normal variables with the correct variance. Prerequisites: PDF, CDF, expectation, variance and affine transformations. Normal distributions are explicitly in GATE 2026 CS. All numerical CDF values used in practice are provided, so the questions do not require a particular table or calculator. Normal approximations are introduced with their limitations, not as an additional syllabus claim.

1. Parameters, density and shape

Write X~N(μ,σ²), where μ is any real mean and σ>0 is standard deviation. The second parameter is variance, not standard deviation. The PDF is

f(x)=[1/(σ√(2π))] exp[−(x−μ)²/(2σ²)], −∞<x<∞.

The density is symmetric about μ, has total area one, and is maximal at μ. Mean, median and unique mode coincide at μ. The variance is σ². Larger σ spreads the curve and lowers its peak while preserving area. The normal model permits every real value; if a physical quantity cannot be negative, a normal approximation may still be useful when the negative tail is negligible, but its support is not an exact physical constraint.

For X~N(50,100), the mean is 50, variance 100 and standard deviation 10. Dividing a deviation by 100 would use the wrong scale. The σ=0 boundary would be a constant variable, not a normal density of the displayed form.

2. Standardization

If X~N(μ,σ²), then Z=(X−μ)/σ has the standard normal distribution N(0,1). Subtracting μ shifts the center to zero; dividing by positive σ rescales spread to one and preserves inequality direction. Define Φ(z)=P(Z≤z).

To find P(X≤x), compute z=(x−μ)/σ and use Φ(z). For a<b,

P(a<X≤b)=Φ((b−μ)/σ)−Φ((a−μ)/σ).

Because normal distributions have densities, each point has probability zero and open/closed endpoints give the same interval probability. The standardized density height φ(z) is not the CDF Φ(z). Most normal interval probabilities have no elementary antiderivative, so CDF tables or numerical values are used.

Example: X~N(50,100). Given Φ(1.5)=.9332, P(X≤65)=Φ((65−50)/10)=.9332. Given Φ(.5)=.6915, P(45<X≤65)=Φ(1.5)−Φ(−.5)=.9332−(1−.6915)=.6247. First compute each z-score; then determine which CDF difference represents the event.

3. Symmetry and one- versus two-sided tails

Symmetry gives Φ(−z)=1−Φ(z). For z≥0, P(Z>z)=1−Φ(z), while P(|Z|>z)=2[1−Φ(z)]. The central interval has probability P(|Z|≤z)=2Φ(z)−1. These formulas use equal tails around zero; they do not apply to an arbitrary asymmetric interval without separate endpoints.

Given Φ(1.96)=.9750, P(Z>1.96)=.0250, P(|Z|>1.96)=.0500 and P(|Z|≤1.96)=.9500. A 95% central region leaves 5% total outside, split as 2.5% in each tail. A one-sided 95th percentile instead uses Φ(z)=.95, approximately z=1.645. Confusing these two z-values changes the event.

The familiar approximate central probabilities within one, two and three standard deviations are 68.27%,95.45%,99.73%. They are normal-model facts, not universal guarantees for all distributions. Do not round “within two standard deviations” to exactly 95% when precise values are supplied.

4. Quantiles reverse the probability calculation

If z_q satisfies Φ(z_q)=q for 0<q<1, then the q-quantile of X is x_q=μ+σz_q. This is the inverse of standardization. Quantiles below the median have negative z-scores, while those above it have positive ones.

Example: X~N(50,100), and z_.95=1.645 is supplied. Its 95th percentile is 50+10(1.645)=66.45. This means P(X≤66.45)≈.95, not that 95% lies within 66.45 units of the mean. For a central 95% interval, use 50±1.96(10), giving approximately [30.4,69.6]. The two tasks use different quantiles.

If a supplied table reports area between zero and positive z rather than the full CDF, add .5 to obtain Φ(z). If it reports upper-tail probability, subtract from one. Always read the table definition; the same displayed number can represent a different area under different conventions.

5. Affine transformations and independent sums

For a≠0, Y=aX+b is normal with mean aμ+b and variance a²σ². Its standard deviation is |a|σ. A negative scale reverses order when solving inequalities, but still gives positive spread. For a=0, Y=b is constant rather than a nondegenerate normal variable.

If X and Y are independent normal variables, then W=aX+bY+c is normal when its variance is positive, with mean aμ_X+bμ_Y+c and variance a²σ_X²+b²σ_Y². Independence supplies both the normality-of-sum result and the zero-covariance simplification in this statement. Merely knowing that both marginal distributions are normal does not guarantee their sum is normal without further joint assumptions.

Example: independent X~N(10,4) and Y~N(20,9). Then X−Y~N(−10,13). The variance is 4+9=13, not 4−9, and the standard deviation is √13, not 2+3. For 2X−Y, the mean is zero and variance 4(4)+9=25. Thus P(2X−Y≤0)=1/2 by symmetry. If X and Y were dependent, an appropriate covariance term would be needed and normality would need separate justification.

6. Optional modeling bridge: normal approximation to counts

A binomial count with sufficiently large np and n(1−p) can often be approximated by N(np,np(1−p)). These conditions are heuristics for approximation quality; no single threshold guarantees every tail accuracy. A discrete count and a continuous normal variable are not identical. The continuity correction represents an integer count k by a unit-width cell around k.

For X~Bin(100,.5), a common approximation for P(X≤60) uses a normal Y with mean 50 and variance 25, and P(Y≤60.5), yielding z=(60.5−50)/5=2.1. Using 60 without correction gives z=2.0 and a different approximation. For P(X≥60), the continuity-corrected boundary is 59.5, not 60.5. Check the inclusive inequality before placing the boundary.

This bridge does not replace exact binomial probabilities when exact answers are required, and it does not claim normality for all sums or sample sizes. Core lesson mastery is standardization, CDF areas, quantiles and the justified independent-normal combination rule.

Checklist

Write down μ, variance and standard deviation separately. Translate the requested event into inequalities before standardizing. Use symmetry to handle negative z and distinguish one tail from two. For quantiles, scale and shift the supplied standard-normal quantile. Add variances, with squared coefficients, only under appropriate assumptions. Label all numerical table rounding and distributional approximations clearly.

Original practice: six single-correct MCQs

These are original questions, not previous-year questions. Choose exactly one option.

NM01. X~N(50,100), where the second parameter is variance. What is the z-score of 65?

A. 0.15 B. 15 C. 1.5 D. 6.5

Answer: C

A: This divides by variance rather than standard deviation.

B: This centers the value but does not scale by the standard deviation.

C: The standard deviation is ten, so (65−50)/10=1.5.

D: This divides by ten without subtracting the mean.

NM02. Z~N(0,1), and Φ(1.96)=0.9750. Find P(|Z|>1.96).

A. 0.05 B. 0.025 C. 0.95 D. 0.975

Answer: A

A: There are two equal tails: 2(1−.975)=.05.

B: This is only the upper tail.

C: This is the central region, the complement of the target.

D: This is P(Z≤1.96), not the two-sided tail.

NM03. X~N(50,100). Given Φ(1.5)=0.9332 and Φ(0.5)=0.6915, find P(45<X≤65).

A. 0.2417 B. 0.9332 C. 0.3085 D. 0.6247

Answer: D

A: This subtracts Φ(.5), but the lower z-score is −.5.

B: This does not subtract the probability below 45.

C: This is the probability below 45 alone.

D: Use .9332−Φ(−.5)=.9332−.3085=.6247.

NM04. X~N(50,100). A standard-normal 95th percentile is 1.645. Find the corresponding percentile of X.

A. 214.5 B. 66.45 C. 69.6 D. 16.45

Answer: B

A: This multiplies the z-quantile by variance rather than standard deviation.

B: Use 50+10×1.645=66.45.

C: This uses 1.96, the upper endpoint for a central 95% interval.

D: This omits the mean shift.

NM05. Independent X~N(10,4) and Y~N(20,9). What is the distribution of X−Y?

A. N(−10,−5) B. N(−10,5) C. N(−10,13) D. N(30,13)

Answer: C

A: Variances do not subtract; a negative variance is impossible.

B: This confuses the sum of standard deviations with variance.

C: Mean subtracts, but independent variances add: −10 and 13.

D: This adds the means instead of subtracting.

NM06. For X~Bin(100,0.5), a continuity-corrected normal approximation to P(X≤60) uses which z-score?

A. 2.0 B. 1.9 C. 0.42 D. 2.1

Answer: D

A: This omits the continuity correction.

B: This uses 59.5, appropriate to the lower boundary for X≥60.

C: This divides by variance 25 rather than standard deviation five.

D: Use boundary 60.5, mean 50 and standard deviation five: 2.1.

Source and scope

Normal distributions are explicitly named in Section 1 Probability and Statistics of the official GATE 2026 CS syllabus. The normal approximation section is a labeled modeling bridge, not a separately claimed syllabus topic. Original questions with supplied CDF values; no full-syllabus completion claim.

Official GATE 2026 CS syllabus (verified 2026-10-14).