CUET UG Applied Mathematics Booster Test 2 - Normal Distribution and Z-Score -

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CUET UG Applied Mathematics Booster Test 2 - 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 2 - Normal Distribution and Z-Score

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QUESTION 1 OF 20

Unlike discrete distributions where outcomes are distinct, a continuous normal distribution is necessary for properties like height or weight because:

Answer Review

1 Unlike discrete distributions where outcomes are distinct, a continuous normal distribution is necessary for properties like height or weight because:

A These properties only take integer values.

B There are an infinite number of values between two distinct values.

C The values can only be negative.

D It is impossible to calculate an average.

💡 Short Explanation

�� Continuous variables can take infinitely many values �� Height and weight are measured on a continuum �� Normal distribution models continuous data

✅ Correct Answer Explanation

Continuous variables such as height and weight can assume infinitely many possible values within an interval. For example, between 50 kg and 51 kg, infinitely many decimal values exist. Thus, a continuous probability distribution like the normal distribution is required. Hence, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Continuous variables are not restricted to integers.
�� Option C → Continuous values can be positive or negative depending on context.
�� Option D → Averages can certainly be calculated for continuous data.

🎯 Strategy

Used

�� Contextual/Tonal Matching

Application:

�� Match the definition of continuous variables with real-world measurements.

Final Logic:

�� Infinite intermediate values define continuity.

🧠 Memory Trick

"Continuous = Infinite Between."

📚 Class XII – NCERT Mathematics Class 12 – Chapter: Probability – Topic: Continuous Random Variables –

2 Match the specific symbols of the normal distribution to their corresponding analytical meaning.

List I	List II
(A) μ	(I) Standard Score
(B) σ	(II) Probability Density Function
(C) Z	(III) Theoretical Mean
(D) f(x)	(IV) Standard Deviation

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

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

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

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

💡 Short Explanation

�� μ denotes mean �� σ denotes standard deviation �� Z represents standardized score

✅ Correct Answer Explanation

Standard notation in normal distribution: μ→ theoretical mean σ→ standard deviation Z→ standard score f(x)→ probability density function Thus: (A) → (III) (B) → (IV) (C) → (I) (D) → (II) Hence, Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Incorrectly swaps mean and standard score.
�� Option C → Misplaces standard deviation and PDF notation.
�� Option D → Incorrectly maps μ to PDF.

🎯 Strategy

Used

�� Option Grouping

Application:

�� Recall standard notation used in probability theory.

Final Logic:

�� Only Option A correctly matches all symbols.

🧠 Memory Trick

"μ Mean, σ Spread, Z Score."

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

3 The normal distribution probability density function f(x) incorporates which of the following mathematical constants and parameters?
(A) Euler's number (e)
(B) Pi (π)
(C) Mean (μ) and Standard Deviation (σ)
(D) Probability of success (p)
Options format:

A A,B,D

B A,B

C A,B,C

D A,B,C,D

💡 Short Explanation

�� Normal PDF contains e and π �� Mean and SD define the distribution �� Probability of success belongs to binomial distribution

✅ Correct Answer Explanation

The normal distribution PDF is: f(x)=1/σ√(2π)e^(-(x-μ)^2/2σ^2) The formula contains: Euler's number e Pi π Mean μ Standard deviation σ The parameter p is used in binomial distributions, not normal PDFs. Hence, Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Incorrectly includes p.
�� Option B → Omits μ and σ, which are essential parameters.
�� Option D → Probability of success p is not part of normal PDF notation.

🎯 Strategy

Used

�� Elimination

Application:

�� Remove all options containing parameter p.

Final Logic:

�� Normal distribution uses e,π,μ,σ.

🧠 Memory Trick

"Normal PDF = e, π, μ, σ."

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

4 Identify the INCORRECT statement regarding the notation and parameters of a normal distribution.

A X is defined as a continuous random variable.

B μ represents the theoretical mean expectation of the distribution.

C σ² denotes the standard deviation directly.

D The notation X ~ N(μ, σ²) indicates the distribution is Gaussian.

💡 Short Explanation

�� σ represents standard deviation �� σ² represents variance �� Normal distributions use notation N(μ,σ^2)

✅ Correct Answer Explanation

In normal distribution notation: X∼N(μ,σ^2) where: μ= mean σ= standard deviation σ^2= variance Thus, Statement C is incorrect because σ^2 denotes variance, not standard deviation. Hence, Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Normal distributions are continuous distributions.
�� Option B → μ correctly represents theoretical mean.
�� Option D → N(μ,σ^2)is standard Gaussian notation.

🎯 Strategy

Used

�� Odd One Out

Application:

�� Identify the statement confusing variance with standard deviation.

Final Logic:

�� σ = SD, while σ² = variance.

🧠 Memory Trick

"Square Means Variance."

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

5 A mixture of normal data points clusters mostly around the center. The probability curve has one peak point, indicating that the normal distribution definitively has:

A Multiple alternating modes

B No definable mode

C A unique mode

D A strictly horizontal linear shape

💡 Short Explanation

�� Normal curve is bell-shaped �� It has one highest point �� Mean = median = mode

✅ Correct Answer Explanation

A normal distribution is unimodal, meaning it has exactly one peak. That peak corresponds to the mode. Since the curve is symmetric: Mean=Median=Mode Thus, the normal distribution has a unique mode. Hence, Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Normal distribution is not multimodal.
�� Option B → The mode exists clearly at the center peak.
�� Option D → Normal curves are bell-shaped, not horizontal.

🎯 Strategy

Used

�� Contextual/Tonal Matching

Application:

�� Match "one peak" with the definition of a unimodal distribution.

Final Logic:

�� One peak implies one unique mode.

🧠 Memory Trick

"Bell Curve = One Peak."

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

6 The continuous constraint region dictates that the normal curve is perfectly symmetrical about the mean μ. Consequently, what is the exact probability area existing strictly ABOVE the mean?

A 1.00

B 0.50

C 0.00

D Infinite

💡 Short Explanation

�� Normal distribution is symmetric about the mean �� Mean divides the curve into two equal halves �� Total area under curve equals 1

✅ Correct Answer Explanation

A normal distribution is perfectly symmetric about its mean μ. Therefore: Half of the total probability lies above the mean Half lies below the mean Since total area under the curve equals 1: P(X>μ)=1/2=0.50 Hence, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Total probability cannot lie entirely above the mean.
�� Option C → Probability above the mean is not zero.
�� Option D → Probability areas are finite and bounded by 1.

🎯 Strategy

Used

�� Contextual/Tonal Matching

Application:

�� Apply symmetry property of the normal distribution.

Final Logic:

�� Symmetry divides total area equally into two halves.

🧠 Memory Trick

"Mean Splits Curve 50–50."

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

7 (Expected Math Index) If the mathematical mean of a normally distributed set of examination scores is precisely 65, what is the statistical value of its median?

A Cannot be determined without variance.

B 0

C 65

D 130

💡 Short Explanation

�� In a normal distribution, mean = median = mode �� Symmetry centers all measures together �� Median equals the mean value

✅ Correct Answer Explanation

A fundamental property of a normal distribution is: Mean=Median=Mode Given: Mean=65 Therefore: Median=65 Hence, Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Variance is unnecessary to determine median here.
�� Option B → Median is not automatically zero unless mean is zero.
�� Option D → There is no basis for doubling the mean.

🎯 Strategy

Used

�� Contextual/Tonal Matching

Application:

�� Use the standard property of the normal distribution.

Final Logic:

�� Mean and median coincide in a symmetric bell curve.

🧠 Memory Trick

"Normal Curve: Mean = Median = Mode."

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

8 The moving average limit of the total area integral from -∞ to +∞ for a normal distribution curve evaluates exactly to 1 unit. What does this unified area statistically represent?

A The standard deviation.

B The Z-score deviation.

C The total combined probability of all possible outcomes.

D The absolute variance.

💡 Short Explanation

�� Total area under PDF equals 1 �� Area represents probability �� Entire sample space probability equals certainty

✅ Correct Answer Explanation

For any probability density function: ∑_(-∞)^∞(f(x) dx=1) This means the total probability of all possible outcomes equals 1. Thus, the total area under the normal curve represents complete certainty that some outcome will occur. Hence, Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Standard deviation measures spread, not total probability.
�� Option B → Z-score measures standardized distance from mean.
�� Option D → Variance measures dispersion, not probability area.

🎯 Strategy

Used

�� Contextual/Tonal Matching

Application:

�� Interpret the meaning of total area under a probability density curve.

Final Logic:

�� Entire area under PDF equals total probability.

🧠 Memory Trick

"Total PDF Area = 1."

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

9 For a standard normal distribution probability model used in Z-score tables, the axis of vertical symmetry occurs precisely at which value?

A Z = -1

B Z = 0

C Z = 1

D Z = σ

💡 Short Explanation

�� Standard normal distribution has mean 0 �� Curve is symmetric about the mean �� Symmetry axis passes through Z=0

✅ Correct Answer Explanation

In the standard normal distribution: μ=0,σ=1 The normal curve is symmetric about its mean. Therefore, the axis of symmetry occurs at: Z=0 Hence, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Z=-1 is one SD below the mean, not the center.
�� Option C → Z=1 lies one SD above the mean.
�� Option D → σ represents spread, not axis location.

🎯 Strategy

Used

�� Elimination

Application:

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

Final Logic:

�� Mean zero defines the symmetry axis.

🧠 Memory Trick

"Standard Normal Centers at Zero."

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

10 A standard normal distribution vector effectively defines its variance Var(X) as exactly 1. What must be the analytical value of the standard deviation σ?

A 0

B 1

C 2

D 0.5

💡 Short Explanation

�� Standard deviation is square root of variance �� Variance of standard normal distribution equals 1 �� Therefore SD also equals 1

✅ Correct Answer Explanation

The relationship between variance and standard deviation is: σ=√(Var(X)) Given: Var(X)=1 Therefore: σ=√(1)=1 Hence, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Standard deviation cannot be zero when variance is 1.
�� Option C → Squaring 2 gives variance 4, not 1.
�� Option D → Squaring 0.5 gives variance 0.25.

🎯 Strategy

Used

�� Substitution

Application:

�� Substitute the variance into the SD formula.

Final Logic:

�� Square root of 1 equals 1.

🧠 Memory Trick

"SD = √Variance."

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

11 To evaluate the comparative area of an IQ score of 90 against a population mean of 100 and a standard deviation of 10, the calculated Standard Score (Z) is:

A 1.0

B -1.0

C 0.1

D -10.0

💡 Short Explanation

�� Use the Z-score formula �� Score is below the mean �� Negative deviation gives negative Z-score

✅ 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% Given: X=90,μ=100,σ=10 Substituting: Z=90-100/10=-10/10=-1 Thus, the score is one standard deviation below the mean. Hence, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Would occur if the score were above the mean by 10.
�� Option C → Incorrect arithmetic calculation.
�� Option D → Ignores division by standard deviation.

🎯 Strategy

Used

�� Substitution

Application:

�� Substitute values directly into the Z-score formula.

Final Logic:

�� (90-100)/10=-1.

🧠 Memory Trick

"Below Mean → Negative Z."

📚 Class XII – NCERT Mathematics Class 12 – Chapter: Probability – Topic: Standard Score (Z-score) –

12 If the integral difference (x - μ) evaluates to exactly zero for a recorded data point, its resulting standard Z-score is mathematically evaluated as:

A 1

B σ

C 0

D Infinite

💡 Short Explanation

�� Z-score depends on deviation from mean �� Zero deviation means score equals mean �� Therefore Z-score becomes zero

✅ Correct Answer Explanation

Using the Z-score formula: Z=x-μ/σ x μ σ z=x-μ/σ≈1.2 Φ(z)≈88.5% If: x-μ=0 then: Z=0/σ=0 Thus, the data point lies exactly at the mean. Hence, Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Z-score equals 1 only when the score is one SD above mean.
�� Option B → σ is standard deviation, not the Z-score value.
�� Option D → Division remains finite because σ is nonzero.

🎯 Strategy

Used

�� Substitution

Application:

�� Directly substitute zero deviation into the formula.

Final Logic:

�� Zero numerator gives Z = 0.

🧠 Memory Trick

"At Mean → Z = 0."

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

📄 Passage

PASSAGE MCQs (COMPULSORY) Calculating the Z-Score is essential for data standardization across diverse data distributions. The fundamental formula for the Z-score is Z = (x - μ) / σ. In an observation study of 7 rare Indian butterflies, the mean length (μ) was calculated as 3 cm with a standard deviation (σ) of 1.69 cm. Notice that butterfly number 3 has a length of 3 cm. With a mean length of 3 cm, its Z-Score evaluates to 0. It means that this data point rests exactly on the mean of the data. Also note that the Z-score is positive if the data point lies above the mean, and negative if it lies below the mean. For example, a butterfly of length 1 cm (which is below the mean) has a negative Z-score of approximately -1.18, effectively standardizing its distance from the center.

According to the passage, a butterfly measured with a length of 5 cm (which is above the mean of 3) will yield a standardized Z-score that is:

A Exactly 0

B Positive

C Negative

D Fractional but strictly less than zero

💡 Short Explanation

�� Scores above mean produce positive Z-values �� Z-score measures direction from mean �� Positive deviation means positive standard score

✅ Correct Answer Explanation

The Z-score formula is: Z=x-μ/σ Given: Butterfly length x=5 cm Mean μ=3 cm Since: x-μ=5-3=2>0 the Z-score must be positive. Hence, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Z-score becomes zero only when x=μ.
�� Option C → Negative Z-scores occur for values below the mean.
�� Option D → A value above the mean cannot produce a negative Z-score.

🎯 Strategy

Used

�� Contextual/Tonal Matching

Application:

�� Interpret whether the value lies above or below the mean.

Final Logic:

�� Above mean implies positive standardized score.

🧠 Memory Trick

"Above Mean → Positive Z."

📚 Class XII – NCERT Mathematics Class 12 – Chapter: Probability – Topic: Positive and Negative Z-Scores –

📄 Passage

Based on the passage, what is the exact formula utilized to achieve data standardization by converting a raw score 'x' into a Z-score?

A (μ - x) / σ

B (x - σ) / μ

C (x - μ) / σ

D x / (μ * σ)

💡 Short Explanation

�� Standardization uses deviation from mean �� Divide deviation by SD �� Produces dimensionless Z-score

✅ Correct Answer Explanation

The standard formula for data standardization is: Z=x-μ/σ x μ σ z=x-μ/σ≈1.2 Φ(z)≈88.5% where: x= raw score μ= mean σ= standard deviation This converts raw data into standardized units. Hence, Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Reverses the sign convention.
�� Option B → Incorrect formula structure.
�� Option D → Does not represent standardization.

🎯 Strategy

Used

�� Memory Recall / Elimination

Application:

�� Recall the standard Z-score equation and eliminate altered forms.

Final Logic:

�� Standardization always uses (x-μ)/σ.

🧠 Memory Trick

"Score minus Mean, over SD."

📚 Class XII – NCERT Mathematics Class 12 – Chapter: Probability – Topic: Z-score Formula –

15 If a set of normally distributed data has a mean of 50 and a standard deviation of 10, the 68.27% probability bracket implies the data will fall roughly between which values?

A 30 and 70

B 40 and 60

C 20 and 80

D 45 and 55

💡 Short Explanation

�� 68.27% data lies within ±1σ �� Mean = 50 and SD = 10 �� Interval becomes 50 ± 10

✅ Correct Answer Explanation

The empirical rule states: 68.27% of data lies within: μ±σ Given: μ=50,σ=10 Therefore: 50-10=40 and 50+10=60 Thus, the interval is 40 to 60. Hence, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Represents approximately ±2σ.
�� Option C → Represents approximately ±3σ.
�� Option D → Too narrow; only ±0.5σ.

🎯 Strategy

Used

�� Substitution

Application:

�� Apply empirical-rule interval directly.

Final Logic:

�� 50±10⇒40 to 60.

🧠 Memory Trick

"68% Lies Within 1σ."

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

16 In a normal distribution, the probability of selecting a data point within the specific range [μ - 2σ, μ + 2σ] is definitively:

A 68.27%

B 95.45%

C 99.73%

D 100.00%

💡 Short Explanation

�� Empirical rule defines standard intervals �� ±2σ contains 95.45% data �� Normal curve probabilities are fixed

✅ Correct Answer Explanation

The empirical rule for a normal distribution states: Within ±1σ → 68.27% Within ±2σ → 95.45% Within ±3σ → 99.73% Therefore: P(μ-2σ<X<μ+2σ)=95.45% Hence, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Corresponds to ±1σ interval.
�� Option C → Corresponds to ±3σ interval.
�� Option D → Total area under entire curve equals 100%, not within ±2σ only.

🎯 Strategy

Used

�� Option Grouping

Application:

�� Match the correct empirical-rule percentage.

Final Logic:

�� ±2σ corresponds to 95.45%.

🧠 Memory Trick

"2σ → 95%."

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

17 Why might a sample consisting of only 25 students' scores be considered statistically suboptimal for conducting a standard Z-test?

A The variance becomes infinitely high.

B The sample size violates the prerequisite of being strictly greater than 30.

C The data points automatically become dependent.

D It mathematically forces multiple modes.

💡 Short Explanation

�� Z-tests generally require large samples �� Standard guideline uses n>30 �� Here sample size is only 25

✅ Correct Answer Explanation

A common condition for applying the Z-test is: n>30 This ensures reliable approximation and stable sampling distribution behavior. Since: n=25<30 the sample size requirement is violated. Hence, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Small sample size does not force infinite variance.
�� Option C → Dependence is unrelated to sample size alone.
�� Option D → Small samples do not automatically create multiple modes.

🎯 Strategy

Used

�� Elimination

Application:

�� Compare sample size with standard Z-test condition.

Final Logic:

�� 25<30, so the large-sample condition fails.

🧠 Memory Trick

"Z-Test → n Above 30."

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

18 During a Z-test preparation, ensuring data is "randomly selected" fundamentally ensures which specific condition is satisfied?

A Sample sizes must inherently be unequal.

B Data points are completely dependent.

C Each data point in the population has an equally likely chance of being selected.

D The standard deviation evaluates to zero.

💡 Short Explanation

�� Random sampling avoids bias �� Every observation gets equal chance �� Independence improves reliability

✅ Correct Answer Explanation

Random selection means each observation in the population has an equal probability of being chosen. This minimizes selection bias and supports independent sampling assumptions used in statistical testing. Hence, Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Random selection does not require unequal sample sizes.
�� Option B → Random selection supports independence, not dependence.
�� Option D → Random sampling does not imply zero standard deviation.

🎯 Strategy

Used

�� Contextual/Tonal Matching

Application:

�� Match the concept of randomness with equal selection opportunity.

Final Logic:

�� Random selection ensures equal chance for all observations.

🧠 Memory Trick

"Random Means Equal Chance."

📚 Class XII – NCERT Mathematics Class 12 – Chapter: Probability – Topic: Random Sampling –

19 For the IQ test scenario where μ = 100 and σ = 10, evaluating the probability P(-1 < Z < 1) yields 0.6826. In raw scores, this represents the probability of scoring precisely between:

A 80 and 120

B 90 and 110

C 95 and 105

D 100 and 110

💡 Short Explanation

�� P(-1<Z<1)means within ±1σ �� Mean = 100 and SD = 10 �� Interval becomes 100 ± 10

✅ Correct Answer Explanation

The interval: -1<Z<1 represents one standard deviation around the mean. Using: X=μ±σ Given: μ=100,σ=10 So: 100-10=90 and 100+10=110 Thus, the interval is 90 to 110. Hence, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Represents ±2σ interval.
�� Option C → Represents only ±0.5σ.
�� Option D → Covers only half the symmetric interval.

🎯 Strategy

Used

�� Substitution

Application:

�� Convert Z-boundaries into raw-score boundaries.

Final Logic:

�� ±1σ around 100 gives 90–110.

🧠 Memory Trick

"One Sigma Around Mean."

📚 Class XII – NCERT Mathematics Class 12 – Chapter: Probability – Topic: Empirical Rule and Z-Scores –

20 In district exam analysis, if Sudha's Z-score is -1.56, this corresponds to a left-tail Z-table area of 0.0594. By multiplying by 100, this concludes her performance was strictly better than what percentage of her batch?

A 94.06%

B 50.00%

C 15.60%

D 5.94%

💡 Short Explanation

�� Left-tail area gives proportion below score �� Convert probability to percentage �� 0.0594×100=5.94%

✅ Correct Answer Explanation

The Z-table gives: P(Z<-1.56)=0.0594 This means 5.94% of students scored below Sudha. Thus, Sudha performed better than: 5.94% of the batch. Hence, Option D is correct.

❓ Why Other Options Are Incorrect

�� Option A → Represents percentage scoring above her, not below.
�� Option B → Would correspond to Z = 0.
�� Option C → Incorrect conversion from probability to percentage.

🎯 Strategy

Used

�� Substitution

Application:

�� Convert decimal probability into percentage directly.

Final Logic:

�� 0.0594×100=5.94%.

🧠 Memory Trick

"Probability × 100 = Percent."

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

CUET UG Applied Mathematics Booster Test 2 - Normal Distribution and Z-Score

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