CUET UG Applied Mathematics Booster Test 3 - Normal Distribution and Z-Score -

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CUET UG Applied Mathematics Booster Test 3 - Normal Distribution and Z-Score

⏱️ 30 Minutes

❓ 20 Questions

📌 Answers are locked once submitted — results and explanations appear at the end.

CUET UG Applied Mathematics Booster Test 3 - Normal Distribution and Z-Score

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QUESTION 1 OF 20

The operational life of a ceiling fan manufactured by Jagdeep Corporation is normally distributed (continuous) with a mean of 300 days and SD of 50 days. To find the probability it lasts "at most 365 days", one must calculate the standard deviate Z. What is this exact Z-score?

Answer Review

1 The operational life of a ceiling fan manufactured by Jagdeep Corporation is normally distributed (continuous) with a mean of 300 days and SD of 50 days. To find the probability it lasts "at most 365 days", one must calculate the standard deviate Z. What is this exact Z-score?

A 1.30

B 1.50

C 1.65

D 2.00

💡 Short Explanation

�� Use the Z-score formula �� Compare observation with mean �� Divide deviation by standard deviation

✅ Correct Answer Explanation

The standard normal deviate is calculated using: Z=X-μ/σ x μ σ z=x-μ/σ≈1.2 Φ(z)≈88.5% Given: X=365,μ=300,σ=50 Substitute values: Z=365-300/50=65/50=1.30 Hence, Option A is correct. Option B incorrectly uses a larger deviation. Option C overestimates the standardized distance. Option D would correspond to a difference of 100 days.

❓ Why Other Options Are Incorrect

�� Option B → Would correspond to a deviation of 75 days from the mean.
�� Option C → Represents a much larger standardized distance.
�� Option D → Implies the observation is 100 days above mean.

🎯 Strategy

Used

�� Substitution

Application:

�� Directly substitute the numerical values into the Z-score formula.

Final Logic:

�� Standardization gives Z=1.30.

🧠 Memory Trick

"Z = Difference ÷ SD."

📚 Class XII – NCERT Mathematics Class 12 – Chapter: Probability – Topic: Standard Normal Distribution –

2 Match the specific Z-score deviation ranges with their corresponding empirical probabilities under the Gaussian distribution curve.

List I	List II
A. -1 to +1	I 0.27%
B. -2 to +2	II 68.27%
C. -3 to +3	III 95.45%
D. Greater than +3 or less than -3	IV 99.73%

A (A)-(II), (B)-(III), (C)-(IV), (D)-(I)

B (A)-(I), (B)-(II), (C)-(III), (D)-(IV)

C (A)-(IV), (B)-(I), (C)-(II), (D)-(III)

D (A)-(III), (B)-(II), (C)-(I), (D)-(IV)

💡 Short Explanation

�� Empirical rule defines standard normal ranges �� ±1σ contains 68.27% data �� Outside ±3σ contains only 0.27% data

✅ Correct Answer Explanation

The empirical rule states: Between -1 and +1: 68.27% Between -2 and +2: 95.45% Between -3 and +3: 99.73% Thus, outside ±3σ: 100%-99.73%=0.27% Hence: (A) → (II) (B) → (III) (C) → (IV) (D) → (I) Therefore, Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Incorrectly swaps the empirical percentages.
�� Option C → Assigns wrong probabilities to standard deviation intervals.
�� Option D → Incorrect mapping of ±1σ and ±3σ regions.

🎯 Strategy

Used

�� Option Grouping

Application:

�� Recall the standard 68–95–99.7 empirical rule.

Final Logic:

�� Only Option A correctly matches all ranges.

🧠 Memory Trick

"68–95–99.7 Rule."

📚 Class XII – NCERT Mathematics Class 12 – Chapter: Probability – Topic: Empirical Rule of Normal Distribution –

3 Let f(x) be the Probability Density Function (PDF) of a continuous normal distribution. Which of the following analytical conditions strictly define f(x)?
(A) f(x) ≥ 0 for all x ∈ (-∞, ∞).
(B) The total integral of f(x) from -∞ to ∞ strictly equals 1.
(C) It utilizes the exponential decay parameter incorporating e^(-(x-μ)²/(2σ²)).
(D) It inherently defines a discrete mathematical function.
Options format:

A A,B,D

B A,B

C A,B,C

D A,B,C,D

💡 Short Explanation

�� PDF values are always non-negative �� Total probability area equals 1 �� Normal PDF contains exponential Gaussian expression

✅ Correct Answer Explanation

For a normal distribution PDF: f(x)=1/σ√(2π)e^(-(x-μ)^2/2σ^2) Thus: A is correct because probabilities cannot be negative. B is correct since total probability equals 1. C is correct because the Gaussian PDF includes the exponential term. Statement D is incorrect because normal distributions are continuous, not discrete. Hence, Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Includes incorrect Statement D.
�� Option B → Omits valid Gaussian exponential condition.
�� Option D → Incorrectly includes discrete-function property.

🎯 Strategy

Used

�� Elimination

Application:

�� Remove all options containing Statement D.

Final Logic:

�� Continuous PDFs cannot be discrete functions.

🧠 Memory Trick

"PDF: Positive, Total 1, Continuous."

📚 Class XII – NCERT Mathematics Class 12 – Chapter: Probability – Topic: Probability Density Function –

4 Identify the INCORRECT relation regarding the structural parameters (μ and σ) of a normal distribution.

A A standard normal distribution is a specifically scaled version where μ = 0 and σ = 1.

B The basic Z-score formula linearly translates raw data x into standard deviation units.

C Changing the mean μ fundamentally alters the spread and width of the bell curve.

D The total geometric area under the curve remains exactly 1 unit regardless of parameter changes.

💡 Short Explanation

�� Mean shifts the curve horizontally �� Standard deviation controls spread �� Total probability area remains constant

✅ Correct Answer Explanation

In a normal distribution: Mean μchanges only the location of the curve. Standard deviation σcontrols spread and width. Thus, Statement C is incorrect because changing μdoes not affect the spread. Statements A, B, and D are correct properties. Hence, Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Correct definition of standard normal distribution.
�� Option B → Z-score converts raw data into SD units.
�� Option D → Total probability under any PDF equals 1.

🎯 Strategy

Used

�� Odd One Out

Application:

�� Identify which statement incorrectly assigns spread behavior to the mean.

Final Logic:

�� Spread depends on σ, not μ.

🧠 Memory Trick

"μ Moves, σ Spreads."

📚 Class XII – NCERT Mathematics Class 12 – Chapter: Probability – Topic: Parameters of Normal Distribution –

5 A mixture of test articles forms a perfect bell-shaped normal curve where 31% of articles are under 45 and 8% are over 64. Because the curve exhibits a unique peak and is symmetric, this data can be utilized via Z-tables directly to solve for the:

A Discrete fractional sum

B Mean μ and Standard Deviation σ

C Poisson event lambda

D Euler's base constant

💡 Short Explanation

�� Normal curves are determined by μ and σ �� Z-table probabilities help estimate these parameters �� Bell-shaped symmetry indicates normal distribution

✅ Correct Answer Explanation

A normal distribution is completely determined by: μandσ Given probabilities and observations, Z-table relationships help calculate these two parameters. Option A is unrelated to continuous distributions. Option C belongs to Poisson distribution. Option D is merely a mathematical constant. Hence, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Fractional sums are unrelated to Gaussian parameter estimation.
�� Option C → Lambda is associated with Poisson distribution.
�� Option D → Euler's constant alone cannot determine a distribution.

🎯 Strategy

Used

�� Contextual/Tonal Matching

Application:

�� Match the bell-shaped distribution concept with its defining parameters.

Final Logic:

�� Every normal distribution is fully defined by μ and σ.

🧠 Memory Trick

"Normal Curve Needs μ and σ."

📚 Class XII – NCERT Mathematics Class 12 – Chapter: Probability – Topic: Parameters of Normal Distribution –

6 Within the symmetrical constraint region of the standardized normal curve, if the area given by P(Z < 1) evaluates to 0.8, what is the precise value of P(Z > -1) due to the curve's perfect symmetry?

A 0.1587

B 0.8

C 0.5000

D 0.6826

💡 Short Explanation

�� Standard normal curve is symmetric about zero �� P(Z<1)=P(Z>-1) �� Equal mirror areas occur on opposite sides

✅ Correct Answer Explanation

For the standard normal distribution: P(Z<1)=0.8 Because the curve is perfectly symmetric about Z=0, P(Z>-1)=P(Z<1) Thus, P(Z>-1)=0.8 Hence, Option B is correct. Option A represents the complementary area beyond Z=1. Option C is only the area on one side of the mean. Option D represents the area between -1 and +1.

❓ Why Other Options Are Incorrect

�� Option A → 0.1587=P(Z>1), not P(Z>-1).
�� Option C → 0.5000 corresponds to half the total area.
�� Option D → Represents central area between -1 and +1.

🎯 Strategy

Used

�� Contextual/Tonal Matching

Application:

�� Apply symmetry property of the standard normal curve.

Final Logic:

�� Mirror-image probabilities are equal in a symmetric distribution.

🧠 Memory Trick

"Mirror Z-values Give Equal Areas."

📚 Class XII – NCERT Mathematics Class 12 – Chapter: Probability – Topic: Symmetry of Standard Normal Distribution –

7 (Expected Math Index) In a perfectly normal IQ dataset, if the mean (EMI) evaluates to exactly 100, what is the strictly defined probability of randomly selecting an individual whose IQ is equal to the median (100) or below?

A 0.50

B 1.00

C 0.00

D 0.68

💡 Short Explanation

�� Normal distribution is symmetric �� Mean = median = mode �� Half the area lies below the mean

✅ Correct Answer Explanation

In a normal distribution: Mean=Median=Mode The distribution is symmetric about the mean. Therefore, exactly half the probability area lies below the mean value. Since the mean IQ is 100: P(X≤100)=0.50 Hence, Option A is correct. Option B would imply certainty. Option C is impossible. Option D refers approximately to probability within one standard deviation.

❓ Why Other Options Are Incorrect

�� Option B → Total probability cannot lie below the mean.
�� Option C → There is always positive probability below the mean.
�� Option D → 0.68 corresponds to the ±1σ interval.

🎯 Strategy

Used

�� Elimination

Application:

�� Use symmetry of the normal curve to remove impossible probability values.

Final Logic:

�� Half the curve lies below the mean.

🧠 Memory Trick

"Mean Splits Curve in Half."

📚 Class XII – NCERT Mathematics Class 12 – Chapter: Probability – Topic: Properties of Normal Distribution –

8 Using a moving average framework over the normal PDF, the formula utilizes a scaling factor of 1 / (σ√(2π)). Mathematically, this constant strictly ensures that:

A The overall variance equates to zero.

B The total geometric area under the entire curve is uniformly scaled to equal 1 unit.

C The distribution smoothly converts into a Poisson model.

D The Z-score translates strictly to negative numbers.

💡 Short Explanation

�� Probability density curves must have total area 1 �� Scaling constant normalizes the curve �� Ensures valid probability distribution

✅ Correct Answer Explanation

The normal distribution PDF is: f(x)=1/σ√(2π)e^(-(x-μ)^2/2σ^2) The factor: 1/σ√(2π) ensures that: ∑_(-∞)^∞(f(x) dx=1) Thus, the total area under the curve equals 1, making it a valid probability density function. Hence, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Variance depends on σ^2, not on the scaling factor alone.
�� Option C → Normal and Poisson distributions are different probability models.
�� Option D → Z-scores can be positive, negative, or zero.

🎯 Strategy

Used

�� Contextual/Tonal Matching

Application:

�� Relate the scaling factor to the normalization condition of PDFs.

Final Logic:

�� The constant ensures total probability equals 1.

🧠 Memory Trick

"PDF Area Always 1."

📚 Class XII – NCERT Mathematics Class 12 – Chapter: Probability – Topic: Probability Density Function –

9 When converting Abhay's performance percentile (better than 44.83%) to a Z-score, the probability area to the left of his score is 0.4483. Because this area is strictly less than 0.5, his Z-score must geometrically lie on which side of the standard mean zero?

A Exactly on zero

B Above zero (positive)

C Below zero (negative)

D Undefined

💡 Short Explanation

�� In standard normal distribution, 0.5 area lies left of mean �� Area less than 0.5 indicates score below mean �� Negative Z-score represents below-average value

✅ Correct Answer Explanation

For the standard normal distribution: P(Z<0)=0.5 Given: P(Z<z)=0.4483 Since 0.4483<0.5, the score lies to the left of the mean. Thus: z<0 Hence, the Z-score is negative. Therefore, Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Z-score equals zero only when area is exactly 0.5.
�� Option B → Positive Z-scores correspond to areas greater than 0.5.
�� Option D → Z-score is clearly determinable from the probability area.

🎯 Strategy

Used

�� Elimination

Application:

�� Compare the cumulative area with 0.5.

Final Logic:

�� Area less than half implies negative Z-score.

🧠 Memory Trick

"Less than 50% → Negative Z."

📚 Class XII – NCERT Mathematics Class 12 – Chapter: Probability – Topic: Interpretation of Z-Scores –

10 A vector of recorded student travel times has μ = 38.8 minutes and σ = 11.4 minutes. A travel time of exactly 50.2 minutes translates to a standard deviate Z-score of:

A Z = 1.0

B Z = 0.0

C Z = -1.0

D Z = 2.0

💡 Short Explanation

�� Use the Z-score formula �� Difference from mean equals one SD �� Positive Z indicates value above mean

✅ Correct Answer Explanation

Use the standard score formula: Z=X-μ/σ x μ σ z=x-μ/σ≈1.2 Φ(z)≈88.5% Substitute values: X=50.2,μ=38.8,σ=11.4Z=50.2-38.8/11.4=11.4/11.4=1.0 Hence, Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Would occur only if X=μ.
�� Option C → Negative Z-score indicates value below mean.
�� Option D → Would require the observation to be two SDs above mean.

🎯 Strategy

Used

�� Substitution

Application:

�� Direct substitution into the Z-score equation gives the result immediately.

Final Logic:

�� One standard deviation above mean gives Z=1.

🧠 Memory Trick

"Difference Equals SD → Z = 1."

📚 Class XII – NCERT Mathematics Class 12 – Chapter: Probability – Topic: Standard Normal Deviate –

11 If a candidate's standard Z-score evaluates to 5, associated with an underlying normal curve mapping where μ = 12 and σ = 4, the original raw data point evaluating to this area is mathematically:

A 20

B 26

C 32

D 40

💡 Short Explanation

�� Use the reverse Z-score formula �� Convert standardized score back to raw score �� Multiply Z by SD and add mean

✅ Correct Answer Explanation

The Z-score formula is: Z=X-μ/σ x μ σ z=x-μ/σ≈1.2 Φ(z)≈88.5% Rearranging: X=μ+Zσ Substitute values: X=12+(5)(4)X=12+20=32 Hence, Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Represents only 12+8, not the required value.
�� Option B → Incorrect arithmetic in reversing the Z-score formula.
�� Option D → Would require Z=7, not Z=5.

🎯 Strategy

Used

�� Substitution

Application:

�� Substitute known Z, mean, and SD into the transformed equation.

Final Logic:

�� Raw score equals 12+20=32.

🧠 Memory Trick

"Raw Score = Mean + Z×SD."

📚 Class XII – NCERT Mathematics Class 12 – Chapter: Probability – Topic: Standard Normal Deviate –

12 If integrating the distance from the mean yields Z = -1.56 for Sudha's score of (where μ = 700 and σ = 180), the specific value -1.56 analytically represents:

A Her raw score numerical difference.

B The number of standard deviations her score rests below the mean.

C Her exact percentile ranking relative to the district.

D The overall calculated variance of her raw score.

💡 Short Explanation

�� Z-score measures distance from mean �� Negative sign indicates below-average score �� Magnitude shows number of SDs away

✅ Correct Answer Explanation

A Z-score tells how many standard deviations a value lies from the mean. Here: Z=-1.56 This means Sudha's score lies: 1.56 standard deviations below the mean. The negative sign indicates the value is less than the mean. Hence, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Z-score is standardized distance, not raw difference.
�� Option C → Percentile requires Z-table lookup, not just the Z-value itself.
�� Option D → Variance is σ^2, unrelated to the Z-score meaning.

🎯 Strategy

Used

�� Contextual/Tonal Matching

Application:

�� Interpret the meaning of a standardized score directly.

Final Logic:

�� Negative Z means "below mean by that many SDs."

🧠 Memory Trick

"Negative Z → Below Mean."

📚 Class XII – NCERT Mathematics Class 12 – Chapter: Probability – Topic: Interpretation of Z-Scores –

📄 Passage

As shown in the normal distribution graph and mathematical theory, if the data values are converted into z-scores within a standard normal distribution, then the percentage of the data that fall within specific numbers of standard deviations (σ) from the mean (μ) for a bell-shaped curve remains remarkably constant. Analytically, there is a 68.27% probability of randomly selecting a Z-score situated strictly between -1 and +1 standard deviations from the mean. Expanding this, there is a 95.45% probability of randomly selecting a score situated between -2 and +2 standard deviations. Finally, there is a 99.73% probability of randomly selecting a score located strictly between -3 and +3 standard deviations from the mean. This is often known statistically as the empirical rule.

Based on the passage, what is the exact evaluated probability of a randomly selected data point falling OUTSIDE the central range of μ - 1σ to μ + 1σ?

A 68.27%

B 31.73%

C 50.00%

D 4.55%

💡 Short Explanation

�� Area within ±1σ equals 68.27% �� Outside area is complementary probability �� Total probability equals 100%

✅ Correct Answer Explanation

From the empirical rule: P(-1<Z<1)=68.27% Therefore, probability outside this region: 100%-68.27%=31.73% Hence, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Represents probability inside ±1σ, not outside.
�� Option C → Incorrect because symmetry does not imply 50% outside ±1σ.
�� Option D → Corresponds approximately to outside ±2σ region.

🎯 Strategy

Used

�� Elimination

Application:

�� Use complementary probability principle.

Final Logic:

�� Outside probability = 1-0.6827=0.3173.

🧠 Memory Trick

"Outside = 100 − Inside."

📚 Class XII – NCERT Mathematics Class 12 – Chapter: Probability – Topic: Empirical Rule –

📄 Passage

Based on the statistical passage, the continuous area representing the absolute probability of randomly selecting a score specifically situated between -2 and +2 standard deviations is numerically formatted as:

A 0.6827

B 0.9545

C 0.9973

D 1.0000

💡 Short Explanation

�� Empirical rule defines standard normal probabilities �� ±2σ contains 95.45% of observations �� Decimal form is 0.9545

✅ Correct Answer Explanation

According to the empirical rule: P(-2<Z<2)=95.45% Converting percentage to decimal: 95.45%=0.9545 Hence, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Represents probability within ±1σ.
�� Option C → Represents probability within ±3σ.
�� Option D → Total probability under entire curve equals 1, not within ±2σ.

🎯 Strategy

Used

�� Option Grouping

Application:

�� Match the correct empirical-rule probability with ±2σ.

Final Logic:

�� ±2σ corresponds to 95.45%.

🧠 Memory Trick

"2σ → 95%."

📚 Class XII – NCERT Mathematics Class 12 – Chapter: Probability – Topic: Standard Normal Distribution –

15 If a researcher maps a normally distributed population but only extracts a sample size of 15 data points, which fundamental Z-test prerequisite is strictly violated?

A Random selection condition

B Equal sample sizes condition

C Sample size greater than 30 condition

D Independent data points condition

💡 Short Explanation

�� Z-tests generally require large samples �� Standard rule uses n>30 �� Here sample size is only 15

✅ Correct Answer Explanation

A major condition for applying the Z-test is: n>30 This ensures reliable approximation to the normal distribution. Since: n=15 the large-sample condition is violated. Hence, Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Random selection is not discussed as violated here.
�� Option B → Equal sample sizes are not required for every Z-test.
�� Option D → Independence is not stated to be violated.

🎯 Strategy

Used

�� Elimination

Application:

�� Compare the sample size with the standard Z-test requirement.

Final Logic:

�� 15<30, so large-sample condition fails.

🧠 Memory Trick

"Z-Test Needs Large n."

📚 Class XII – NCERT Mathematics Class 12 – Chapter: Probability – Topic: Z-Test Fundamentals –

16 If an academic investigator purposely chooses only the highest performing students for a Z-test evaluation, this inherently violates the prerequisite that all data points must be:

A Correlated directly to the mean.

B Randomly selected and independent.

C Strictly greater than a quantity of 30.

D Binomial by underlying nature.

💡 Short Explanation

�� Z-tests require unbiased sampling �� Selecting only top performers creates bias �� Data must be random and independent

✅ Correct Answer Explanation

A fundamental assumption of the Z-test is that observations are: Randomly selected Independent of each other Choosing only the highest-performing students introduces selection bias and destroys randomness. Hence, the prerequisite violated is random and independent sampling. Therefore, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Correlation with mean is not a Z-test requirement.
�� Option C → Sample size condition is unrelated to selective bias here.
�� Option D → Z-tests do not require binomial data specifically.

🎯 Strategy

Used

�� Contextual/Tonal Matching

Application:

�� Match "purposely choosing highest performers" with lack of randomness.

Final Logic:

�� Biased selection violates random independent sampling.

🧠 Memory Trick

"Z-Test Needs Random Data."

📚 Class XII – NCERT Mathematics Class 12 – Chapter: Probability – Topic: Conditions for Statistical Testing –

17 For the standard IQ distribution analysis (μ = 100, σ = 10), calculating P(X < 90) requires evaluating P(Z < -1). If the symmetric table area P(Z < 1) evaluates to 0.8, what is the exact probability P(Z < -1)?

A 0.1587

B 0.3174

C 0.5000

D 0.8

💡 Short Explanation

�� Standard normal curve is symmetric �� Left tail probability below -1 equals right tail beyond +1 �� Use complement of cumulative area

✅ Correct Answer Explanation

Given: P(Z<1)=0.8 Therefore, P(Z>1)=1-0.8=0.1587 By symmetry of the normal distribution: P(Z<-1)=P(Z>1)=0.1587 Hence, Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Represents probability outside ±1 together.
�� Option C → Corresponds to area left of mean Z=0.
�� Option D → Represents probability below Z=1, not -1.

🎯 Strategy

Used

�� Elimination

Application:

�� Use complement and symmetry properties of the standard normal curve.

Final Logic:

�� P(Z<-1)=1-0.8=0.1587.

🧠 Memory Trick

"Tail Areas Mirror."

📚 Class XII – NCERT Mathematics Class 12 – Chapter: Probability – Topic: Standard Normal Probabilities –

18 Volunteer A scored 74 on an IQ test where μ = 62 and σ = 11. His standardized Z-score is roughly 1.09. From the Z-table, the cumulative left-area for Z = 1.09 is 0.8621. Statistically, this proves Volunteer A performed better than approximately what percentage of volunteers?

A 13.79%

B 86.21%

C 10.90%

D 50.00%

💡 Short Explanation

�� Cumulative Z-table area gives proportion below score �� 0.8621 means 86.21% scored lower �� Higher percentile means better performance

✅ Correct Answer Explanation

The cumulative probability: P(Z<1.09)=0.8621 means 86.21% of observations lie below Volunteer A's score. Therefore, Volunteer A performed better than approximately: 86.21% of volunteers. Hence, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Represents only the upper tail area.
�� Option C → Confuses Z-score value with percentile.
�� Option D → Would correspond to average performance.

🎯 Strategy

Used

�� Contextual/Tonal Matching

Application:

�� Interpret cumulative probability directly as percentile rank.

Final Logic:

�� Area to the left equals percentage scoring below him.

🧠 Memory Trick

"Left Area = Percent Below."

📚 Class XII – NCERT Mathematics Class 12 – Chapter: Probability – Topic: Interpretation of Z-Tables –

19 If a student's examination score mathematically transforms to a Z-score of exactly 0.00, it strictly analytical implies that the raw score is exactly equal to:

A The standard deviation σ.

B The overall variance σ².

C The theoretical mean expectation μ.

D Absolute zero.

💡 Short Explanation

�� Z-score measures distance from mean �� Z = 0 means no deviation from mean �� Raw score equals mean exactly

✅ Correct Answer Explanation

The Z-score formula is: Z=X-μ/σ x μ σ z=x-μ/σ≈1.2 Φ(z)≈88.5% If: Z=0 then: X-μ=0 Thus: X=μ Hence, the raw score equals the mean. Therefore, Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Standard deviation measures spread, not central value.
�� Option B → Variance is spread squared, unrelated to equality condition.
�� Option D → Absolute zero is unrelated to Z-scores.

🎯 Strategy

Used

�� Substitution

Application:

�� Substitute Z=0 into the standardization formula.

Final Logic:

�� Zero standardized distance means the value equals the mean.

🧠 Memory Trick

"Z = 0 → At Mean."

📚 Class XII – NCERT Mathematics Class 12 – Chapter: Probability – Topic: Z-Score Interpretation –

20 Data standardization utilizing the Z-score equation Z = (x - μ) / σ basically performs a linear transformation on a dataset. This shifts the original curve to center strictly at zero and scales the width so that the new variance invariably equals:

A 0

B 1

C σ²

D μ

💡 Short Explanation

�� Standardization creates standard normal distribution �� Mean becomes 0 �� Variance becomes 1

✅ Correct Answer Explanation

Standardization transforms data using: Z=X-μ/σ x μ σ z=x-μ/σ≈1.2 Φ(z)≈88.5% After transformation: Mean becomes: Standard deviation becomes: 1 Therefore, variance becomes: 1^2=1 Hence, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Variance cannot become zero after standardization.
�� Option C → Original variance changes during standardization.
�� Option D → Mean becomes zero, not μ.

🎯 Strategy

Used

�� Contextual/Tonal Matching

Application:

�� Recall the defining properties of the standard normal distribution.

Final Logic:

�� Standardization always produces variance equal to 1.

🧠 Memory Trick

"Standard Normal → Mean 0, Variance 1."

📚 Class XII – NCERT Mathematics Class 12 – Chapter: Probability – Topic: Standardization of Data –

CUET UG Applied Mathematics Booster Test 3 – Normal Distribution and Z-Score

CUET UG Applied Mathematics Booster Test 3 - Normal Distribution and Z-Score

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