Let the Mean and Variance of five observations x1 = 1, x2 = 3, x3 = a, x4 = 7 and x5 = b, a > b, be 5 and 10 respectively. Then the Variance of the observations n + xn, n = 1, 2, 5 is

Mean of N observations: Variance of N observations: Given mean is 5 for 5 observations: Given variance is 10: Using standard algebraic identity (a+b)2 = a2 + b2 + 2ab: Solving a+b=14 and ab=40: Since a > b, we obtain a = 10 and b = 4. The original set is x1 = 1, x2 = 3, x3 = 10, x4 = 7, x5 = 4. We construct the new set yn = n + xn: - y1 = 1 + 1 = 2 - y2 = 2 + 3 = 5 - y3 = 3 + 10 = 13 - y4 = 4 + 7 = 11 - y5 = 5 + 4 = 9 Mean of new set: Variance of new set: Note that adding a changing factor like +n is different from adding a constant C to each observation (which leaves variance unchanged). In this case, calculate individual xn variables directly first before applying transformations. Class 11 Mathematics: Statistics and Probability

Let the Mean and Variance of five observations x1 = 1, x2 = 3, x3 = a, x4 = 7 and x5 = b, a > b, be 5 and 10 respectively. Then the Variance of the observations n + xn, n = 1, 2, 5 is

Mean of N observations: Variance of N observations: Given mean is 5 for 5 observations: Given variance is 10: Using standard algebraic identity (a+b)2 = a2 + b2 + 2ab: Solving a+b=14 and ab=40: Since a > b, we obtain a = 10 and b = 4. The original set is x1 = 1, x2 = 3, x3 = 10, x4 = 7, x5 = 4. We construct the new set yn = n + xn: - y1 = 1 + 1 = 2 - y2 = 2 + 3 = 5 - y3 = 3 + 10 = 13 - y4 = 4 + 7 = 11 - y5 = 5 + 4 = 9 Mean of new set: Variance of new set: Note that adding a changing factor like

Let the Mean and Variance of five observations x1 = 1, x2 = 3, x3 = a, x4 = 7 and x5 = b, a > b, be 5 and 10 respectively. Then the Variance of the observations n + xn, n = 1, 2, 5 is

Mean of N observations: Variance of N observations: Given mean is 5 for 5 observations: Given variance is 10: Using standard algebraic identity (a+b)2 = a2 + b2 + 2ab: Solving a+b=14 and ab=40: Since a > b, we obtain a = 10 and b = 4. The original set is x1 = 1, x2 = 3, x3 = 10, x4 = 7, x5 = 4. We construct the new set yn = n + xn: - y1 = 1 + 1 = 2 - y2 = 2 + 3 = 5 - y3 = 3 + 10 = 13 - y4 = 4 + 7 = 11 - y5 = 5 + 4 = 9 Mean of new set: Variance of new set: Note that adding a changing factor like +n is different from adding a constant C to each observation (which leaves variance unchanged). In this case, calculate individual xn variables directly first before applying transformations. Class 11 Mathematics: Statistics and Probability

Let the Mean and Variance of five observations x1 = 1, x2 = 3, x3 = a, x4 = 7 and x5 = b, a > b, be 5 and 10 respectively. Then the Variance of the observations n + xn, n = 1, 2, 5 is

Mean of N observations: Variance of N observations: Given mean is 5 for 5 observations: Given variance is 10: Using standard algebraic identity (a+b)2 = a2 + b2 + 2ab: Solving a+b=14 and ab=40: Since a > b, we obtain a = 10 and b = 4. The original set is x1 = 1, x2 = 3, x3 = 10, x4 = 7, x5 = 4. We construct the new set yn = n + xn: - y1 = 1 + 1 = 2 - y2 = 2 + 3 = 5 - y3 = 3 + 10 = 13 - y4 = 4 + 7 = 11 - y5 = 5 + 4 = 9 Mean of new set: Variance of new set: Note that adding a changing factor like

Let the Mean and Variance of five observations x1 = 1, x2 = 3, x3 = a, x4 = 7 and x5 = b, a

Solution & Explanation

Related Formula

Mean of $N$ $N$ observations:

x = (Σ xᵢ)/(N)

Variance of $N$ $N$ observations:

σ² = (Σ xᵢ²)/(N) - ( x)²

Core Logic

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

(1 + 3 + a + 7 + b)/(5) = 5 a + b + 11 = 25 a + b = 14 --- (1)

Given variance is $10$ $10$:

(1² + 3² + a² + 7² + b²)/(5) - 5² = 10 (59 + a² + b²)/(5) = 35

a² + b² = 175 - 59 = 116 --- (2)

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

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

14² = 116 + 2ab 196 = 116 + 2ab ab = 40

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

a(14-a) = 40 a² - 14a + 40 = 0 (a-10)(a-4) = 0

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 $x₁ = 1, x₂ = 3, x₃ = 10, x₄ = 7, x₅ = 4$ . 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:

y = (2 + 5 + 13 + 11 + 9)/(5) = (40)/(5) = 8

Variance of new set:

σnew² = (2² + 5² + 13² + 11² + 9²)/(5) - 8²

σnew² = (4 + 25 + 169 + 121 + 81)/(5) - 64 = (400)/(5) - 64 = 80 - 64 = 16

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

More Statistics and Probability Previous-Year Questions — Page 5

Q12 jee_main_2024_31_jan_evening Mean and Variance

Let the mean and the variance of 6 observation

a, b, 68, 44, 48, 60

55

and

194

, respectively if

a > b

, then

a + 3b

A. $200$ $200$
B. $190$ $190$
C. $180$ $180$
D. $210$ $210$

Solution

Related Formula

Mean x = (Σ xᵢ)/(n)

Variance σ² = Σ (xᵢ - x)²n

Core Logic

Mean is 55:

(a + b + 68 + 44 + 48 + 60)/(6) = 55

220 + a + b = 330 a + b = 110

Variance is 194:

((a-55)² + (b-55)² + (68-55)² + (44-55)² + (48-55)² + (60-55)²)/(6) = 194

(a-55)² + (b-55)² + 13² + (-11)² + (-7)² + 5² = 1164

(a-55)² + (b-55)² + 169 + 121 + 49 + 25 = 1164

(a-55)² + (b-55)² = 800

Expand the squares using $a+b=110$ $a+b=110$: $a² + b² - 110(a+b) + 2(3025) = 800$ $a^2 + b^2 - 110(a+b) + 2(3025) = 800$ $a² + b² - 110(110) + 6050 = 800 a² + b² = 6850$ Using $(a+b)² = 12100 a²+b²+2ab = 12100 2ab = 12100 - 6850 = 5250$ . $(a-b)² = a²+b² - 2ab = 6850 - 5250 = 1600 a-b = 40$ (since $a>b$ $a>b$).

Solving $a+b=110$ $a+b=110$ and $a-b=40$ $a-b=40$:

a = 75, b = 35

Finally, evaluate $a + 3b$ $a + 3b$:

a + 3b = 75 + 3(35) = 180

Chapter Mix

Class 11 Maths: Statistics

Solution & Explanation

Related Formula

Core Logic

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

Step 2: Constructing new set and finding variance

Pattern Recognition

Chapter Mix

Reference Study Guides

More Statistics and Probability Previous-Year Questions — Page 5

Solution

Related Formula

Core Logic

Chapter Mix

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Let the Mean and Variance of five observations x₁ = 1, x₂ = 3, x₃ = a, x₄ = 7$x_1 = 1, x_2 = 3, x_3 = a, x_4 = 7$ and x₅ = b, a > b,$x_5 = b, a > b,$ be 5$5$ and 10$10$ respectively. Then the Variance of the observations n + xₙ, n = 1, 2, 5$n + x_n, n = 1, 2, \dots \dots 5$ is

Solution & Explanation

Related Formula

Core Logic

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

Step 2: Constructing new set and finding variance

Pattern Recognition

Chapter Mix

Reference Study Guides

More Statistics and Probability Previous-Year Questions — Page 5

Solution

Related Formula

Core Logic

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