Statistics Puzzle: Variance under Linear Transformation

Given:

• Y = 10.5 + 2X
• Var(Y) = 100

Formula: If Y = a + bX, then Var(Y) = b² × Var(X)

Apply:

• 100 = 2² × Var(X)
• 100 = 4 × Var(X)
• Var(X) = 100 / 4 = 25

✅ Final Answer: The variance of the marks given by the first judge is 25.

Given two sets:

• Set \( A \) has median \( m_1 = 2 \)
• Set \( B \) has median \( m_2 = 4 \)

What can we say about the median of the combined set \( A \cup B \)?

✅ Answer:

The combined median depends on the size and values of both sets.

Without that information, we only know that:

\[ \text{Combined Median} \in [2, 4] \]

So, the exact median cannot be determined with the given data.

Mean, Median, and Mode Relation

Given:

• Mean = 1
• Median = \( 3^x \)
• Mode = \( 9^x \)

Use empirical formula:

\[ \text{Mode} = 3 \cdot \text{Median} - 2 \cdot \text{Mean} \]

\[ 9^x = 3 \cdot 3^x - 2 \Rightarrow (3^x)^2 = 3 \cdot 3^x - 2 \]

Let \( y = 3^x \), then:

\[ y^2 = 3y - 2 \Rightarrow y^2 - 3y + 2 = 0 \Rightarrow (y - 1)(y - 2) = 0 \]

So, \( y = 1 \text{ or } 2 \Rightarrow 9^x = y^2 = 1 \text{ or } 4 \)

✅ Final Answer: \(\boxed{1 \text{ or } 4}\)

Given: Mean of squares of first n natural numbers is 11.

Use the identity for sum of squares: $$\sum_{k=1}^{n} k^2=\frac{n(n+1)(2n+1)}{6}$$

Mean \(=\) (sum) \( \div n \Rightarrow \) $$\frac{1}{n}\sum_{k=1}^{n}k^2=\frac{(n+1)(2n+1)}{6}=11$$

Solve for \(n\): $$(n+1)(2n+1)=66 $$ $$\;\Rightarrow\; 2n^2+3n+1=66 \;\Rightarrow\; $$ $$n=\frac{-3\pm\sqrt{9+520}}{4}=\frac{-3\pm 23}{4}$$

Only positive integer solution: $$n=\frac{20}{4}=5$$

Answer: \(n=5\).

Concept: Standard deviation does not change when the same constant is added to every observation.

For data \(x_i\) and constant \(k\): $$\operatorname{SD}(x_i+k)=\operatorname{SD}(x_i).$$

Given: Original SD = 30, each number increased by 4.

New standard deviation: \(30\).

Question: Which one of the following is NOT a correct statement?

1. The value of standard deviation changes by a change of scale.
2. The standard deviation is greater than or equal to the mean deviation (about mean).
3. The sum of squares of deviations is minimum when taken from the mean.
4. The variance is expressed in the same units as the units of observation.

Answer: Option 4

Why: Variance has squared units, not the same units as the data (e.g., if data are in cm, variance is in cm²). Standard deviation (the square root of variance) has the same units as the data.

Notes:

• Change of scale: if each value is multiplied by \(k\), then \(\text{SD}\) becomes \(|k|\cdot \text{SD}\) and \(\text{Var}\) becomes \(k^2\cdot \text{Var}\).
• \(\text{SD} \ge \text{Mean Deviation (about mean)}\) is a standard inequality.
• \(\sum (x_i - a)^2\) is minimized at \(a = \bar{x}\) (the mean).

Given: Values \(0,1,2,\dots,50\) with frequencies \(1,\binom{50}{1},\binom{50}{2},\dots,\binom{50}{50}\) (i.e., \(\binom{50}{k}\) for value \(k\)).

Total frequency: $$\sum_{k=0}^{50}\binom{50}{k}=2^{50}.$$

Sum of (value × frequency): $$\sum_{k=0}^{50} k\binom{50}{k}=50\cdot 2^{49}\quad\text{(identity: }\sum k\binom{n}{k}=n2^{\,n-1}\text{)}.$$

Arithmetic Mean (AM): $$\bar{x}=\frac{\sum k\binom{50}{k}}{\sum \binom{50}{k}} =\frac{50\cdot 2^{49}}{2^{50}}=\frac{50}{2}=25.$$

Answer: \(25\).

Given: \(n=200\), reported mean \(=40\) ⇒ reported total $$S_{\text{reported}}=200\times 40=8000.$$

Two scores were misread: true \((43,35)\) but used as \((34,53)\).

Correct the total: $$S_{\text{correct}}=S_{\text{reported}}-\big(34+53\big)+\big(43+35\big) =8000-87+78=7991.$$

Corrected mean: $$\bar{x}_{\text{correct}}=\frac{S_{\text{correct}}}{n} =\frac{7991}{200}=39.955\;\approx\;39.96.$$

Answer: \(39.955\) (≈ \(39.96\)).

AP: \(a,\,a+d,\,a+2d,\,\dots,\,a+2nd\) has \(2n+1\) terms. The mean is the middle term \(a+nd\).

Mean deviation from the mean (MD): average of absolute deviations from \(a+nd\).

Symmetry gives pairs at distances \(kd\) for \(k=1,\dots,n\). Hence

Answer: \(\displaystyle \boxed{\frac{d\,n(n+1)}{2n+1}}\).

Given: Females = 40 (avg = 5.15 ft), Males = 20 (avg = 5.66 ft). Total students = 60.

Weighted average height = $$\frac{40\times 5.15 + 20\times 5.66}{40+20}$$

Compute: $$\frac{206 + 113.2}{60}=\frac{319.2}{60}=5.32$$

Answer: 5.32 feet

Given AP: \(1,\,1+d,\,1+2d,\,\ldots,\,1+100d\) (total \(101\) terms). The mean is the middle term \(1+50d\).

Mean Deviation (about mean): For \(2n+1\) terms \(a,a+d,\ldots,a+2nd\), $$\text{MD}=\frac{d\,n(n+1)}{2n+1}.$$

Here \(2n+1=101 \Rightarrow n=50\). So $$\text{MD}=\frac{d\cdot 50\cdot 51}{101}=\frac{2550}{101}\,d.$$

Given \(\text{MD}=255\). Solve for \(d\): $$\frac{2550}{101}\,d=255 \;\Rightarrow\; d=\frac{255\times 101}{2550}=10.1.$$

Answer: \(d=10.1\).

Given sets

• A: 2, 3, 7, 1, 3, 2, 3
• B: 7, 5, 9, 12, 5, 3, 8
• C: 4, 4, 11, 7, 2, 3, 4

Set A: sum = 21, n = 7 ⇒ Mean = 21/7 = 3.
Sorted A = (1, 2, 2, 3, 3, 3, 7) ⇒ Median = 3.
Mode(A) = most frequent = 3.

Set B: sorted = (3, 5, 5, 7, 8, 9, 12) ⇒ Median(B) = 7.

Set C: sum = 35, n = 7 ⇒ Mean(C) = 5; Mode(C) = 4.

Check options

1. Mean(A) = 3 vs Mode(C) = 4 → False
2. Mean(C) = 5 vs Median(B) = 7 → False
3. Median(B) = 7 vs Mode(A) = 3 → False
4. Mean(A) = Median(A) = Mode(A) = 3 → True

Correct statement: Option 4

Corrected Mean and Standard Deviation

Original Mean: 40, Standard Deviation: 15

Two scores were misread: 25 → 52 and 35 → 53

Corrected Mean:

\( \mu' = \frac{3955}{100} = \boxed{39.55} \)

Corrected Standard Deviation:

\( \sigma' = \sqrt{\frac{178837}{100} - (39.55)^2} \approx \boxed{14.96} \)

✅ Final Answer: Mean = 39.55, Standard Deviation ≈ 14.96

Median & Frequency Table

Given: Median = 42.6, Total Frequency = 685

Using Median Formula:

\( \text{Median} = L + \left( \frac{N/2 - F}{f} \right) \cdot h \)

• Median Class: 40–50
• Lower boundary \( L = 40 \)
• Frequency \( f = 180 \)
• Class width \( h = 10 \)
• Cumulative freq before median class \( F = 214 + f_1 \)

Substituting values:

\( 42.6 = 40 + \left( \frac{128.5 - f_1}{180} \right) \cdot 10 \Rightarrow f_1 = \boxed{82} \)

Using total frequency:

\( 662 + f_2 = 685 \Rightarrow f_2 = \boxed{23} \)

✅ Final Answer: \( f_1 = 82,\quad f_2 = 23 \)

Given averages: Boys = 52, Girls = 42, Combined = 50.

Let the proportion of boys be \(p\) (so girls = \(1-p\)). Weighted mean gives: $$52p + 42(1-p) = 50$$ $$52p + 42 - 42p = 50 \Rightarrow 10p = 8 \Rightarrow p = 0.8.$$

Percentage of boys \(= 0.8 \times 100\% = \boxed{80\%}\).