If the mean and the variance of the data

Class	4–8	8–12	12–16	16–20
Frequency	3	λ$\lambda$	4	7

are μ$\mu$ and 19 respectively, then the value of λ + μ$\lambda + \mu$ is

Related Formula

Core Logic

For X = 1, 2, , 19$X = \{1, 2, \dots, 19\}$, we find the mean x$\bar{x}$ and variance σₓ²$\sigma_x^2$.

Step 1: Creating System of Equations

Given Mean of Y = 30$Y = 30$ and Variance of Y = 750$Y = 750$.

Using transformations:

Step 2: Solving for a and b

From the variance equation:

Now find the corresponding values of b$b$ from equation (1): If a = 5 b = 30 - 10(5) = 30 - 50 = -20$a = 5 \implies b = 30 - 10(5) = 30 - 50 = -20$ If a = -5 b = 30 - 10(-5) = 30 + 50 = 80$a = -5 \implies b = 30 - 10(-5) = 30 + 50 = 80$

Step 3: Finding Sum of b Values

The possible values for b$b$ are -20$-20$ and 80$80$.

Sum = -20 + 80 = 60$= -20 + 80 = 60$

Pattern Recognition

When a dataset undergoes linear transformation Y = aX + b$Y = aX + b$, the variance isolates a²$a^2$ entirely decoupled from b$b$. Solve variance first to get a$a$, then back-substitute into the mean equation.

Chapter Mix

Class 11 Maths: Statistics

Related Formula

Mean x = (Σ xᵢ)/(n)$\bar{x} = \frac{\sum x_i}{n}$ Variance σ² = (Σ xᵢ²)/(n) - ( x)²$\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$ Mean Deviation about Median = (Σ |xᵢ - M|)/(n)$\frac{\sum |x_i - M|}{n}$ where M$M$ is the median.

Core Logic

Let the two missing observations be a$a$ and b$b$. From the given mean:

From the given variance:

Step 1: Solve for a and b

Sum of squares of knowns:

a² + b² = 58$a^2 + b^2 = 58$

We know a+b = 10 b = 10-a$a+b = 10 \implies b = 10-a$.

Thus, the missing observations are 3 and 7.

Step 2: Find the Median

Arrange all 10 observations in ascending order: 2, 3, 3, 5, 7, 10, 11, 13, 15, 21$2, 3, 3, 5, 7, 10, 11, 13, 15, 21$ Since there are 10 observations, median M$M$ is the average of the 5th and 6th terms:

Step 3: Calculate Mean Deviation about Median

Find absolute deviations |xᵢ - M|$|x_i - M|$: |2-8.5|=6.5$|2-8.5|=6.5$ |3-8.5|=5.5$|3-8.5|=5.5$ |3-8.5|=5.5$|3-8.5|=5.5$ |5-8.5|=3.5$|5-8.5|=3.5$ |7-8.5|=1.5$|7-8.5|=1.5$ |10-8.5|=1.5$|10-8.5|=1.5$ |11-8.5|=2.5$|11-8.5|=2.5$ |13-8.5|=4.5$|13-8.5|=4.5$ |15-8.5|=6.5$|15-8.5|=6.5$ |21-8.5|=12.5$|21-8.5|=12.5$

Sum of absolute deviations:

Chapter Mix

Class 11 Mathematics: Statistics

Related Formula

Core Logic

We set up algebraic equations using the definitions of mean and variance to determine the values of a+b$a+b$ and ab$ab$.

Step 1: Apply the Mean condition

Given mean x = 9$\bar{x} = 9$ for N=8$N=8$ observations:

Step 2: Apply the Variance condition

Given variance σ² = 9.25 = (37)/(4)$\sigma^2 = 9.25 = \frac{37}{4}$:

Step 3: Solve for ab and calculate the target expression

We know (a+b)² = a² + b² + 2ab$(a+b)^2 = a^2 + b^2 + 2ab$. Substitute equations (1) and (2):

Now, calculate the target value:

Pattern Recognition

Direct symmetric evaluation: Statistics problems in JEE with missing observations often ask for symmetric combinations of variables like a+b+ab$a+b+ab$ or a²+b²$a^2+b^2$. These can be calculated using quadratic expansions without solving for a$a$ and b$b$ individually.

Chapter Mix

Class 11 Mathematics: Statistics

Related Formula

Mean of N$N$ observations:

Variance of N$N$ observations:

Core Logic

Given mean is 5$5$ for 5$5$ observations:

Given variance is 10$10$:

Step 1: Finding a$a$ and b$b$

Using standard algebraic identity (a+b)² = a² + b² + 2ab$(a+b)^2 = a^2 + b^2 + 2ab$:

Solving a+b=14$a+b=14$ and ab=40$ab=40$:

Since a > b$a > b$, we obtain a = 10$a = 10$ and b = 4$b = 4$.

Step 2: Constructing new set and finding variance

The original set is x₁ = 1, x₂ = 3, x₃ = 10, x₄ = 7, x₅ = 4$x_1 = 1, x_2 = 3, x_3 = 10, x_4 = 7, x_5 = 4$. We construct the new set yₙ = n + xₙ$y_n = n + x_n$:

• y₁ = 1 + 1 = 2$y_1 = 1 + 1 = 2$
• y₂ = 2 + 3 = 5$y_2 = 2 + 3 = 5$
• y₃ = 3 + 10 = 13$y_3 = 3 + 10 = 13$
• y₄ = 4 + 7 = 11$y_4 = 4 + 7 = 11$
• y₅ = 5 + 4 = 9$y_5 = 5 + 4 = 9$

Mean of new set:

Variance of new set:

Pattern Recognition

Note that adding a changing factor like +n$+n$ is different from adding a constant C$C$ to each observation (which leaves variance unchanged). In this case, calculate individual xₙ$x_n$ variables directly first before applying transformations.

Chapter Mix

Class 11 Mathematics: Statistics and Probability

Related Formula

Sum of all probabilities in a distribution:

Complementary probability:

Core Logic

Let's first determine the constant k$k$:

This is an Arithmetico-Geometric Progression (AGP). Let S = Σx=0∞ (x+1)((1)/(3))^x$S = \sum_{x=0}^{\infty} (x+1)\left(\frac{1}{3}\right)^x$:

Step 1: Finding k$k$

Subtracting (2) from (1):

Substitute back:

Step 2: Calculating P(X ≥ 3)$P(X \ge 3)$

Calculate initial probability values:

• P(X=0) = k(1)(1) = (4)/(9)$P(X=0) = k(1)(1) = \frac{4}{9}$
• P(X=1) = k(2)((1)/(3)) = (2)/(3) · (4)/(9) = (8)/(27)$P(X=1) = k(2)\left(\frac{1}{3}\right) = \frac{2}{3} \cdot \frac{4}{9} = \frac{8}{27}$
• P(X=2) = k(3)((1)/(9)) = (1)/(3) · (4)/(9) = (4)/(27)$P(X=2) = k(3)\left(\frac{1}{9}\right) = \frac{1}{3} \cdot \frac{4}{9} = \frac{4}{27}$

Pattern Recognition

The infinite AGP sum with factor (x+1)r^x$(x+1)r^x$ always converges to (1)/((1-r)²)$\frac{1}{(1-r)^2}$. Here r = 1/3$r = 1/3$, so the sum is (1)/((2/3)²) = 9/4$\frac{1}{(2/3)^2} = 9/4$. This mental check saves doing the full subtraction sequence.

Chapter Mix

Class 11 Mathematics: Statistics and Probability Class 11 Mathematics: Sequences and Series