The number of values of <math> <mrow> <mi>a</mi> <mo>∈</mo> <mi>N</mi> </mrow> </math> such that the variance of 3, 7, 12, a, 43 -a is a natural number is:

M e a n = 3 + 1 2 + 7 + a + ( 4 3 − a ) 5 = 1 3 Variance = 3 2 + 1 2 2 + 7 2 + a 2 + ( 4 3 − a ) 2 5 − ( 1 3 ) 2 ⇒ 2 a 2 − a + 1 5 → N a t u r a l n u m b e r Let 2a2 – a + 1 = 5xD = 1 – 4 (2) (1 – 5n)= 40n – 7, which is not 4 λ o r 4 λ + 1 f r o m . As each square form is 4 λ o r 4 λ + 1

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Variance Maths Statistics Maths Class 11th

The number of values of $a \in N$ such that the variance of 3, 7, 12, a, 43 -a is a natural number is:

Option 1 -

Option 2 -

Option 3 -

Option 4 -

infinite

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alok kumar singh | Contributor-Level 10

2 weeks ago

Correct Option - 1

Detailed Solution:

$M e a n = \frac{3 + 12 + 7 + a + (43 - a)}{5} = 13$

Variance = $\frac{3^{2} + 1 2^{2} + 7^{2} + a^{2} + {(43 - a)}^{2}}{5} - {(13)}^{2}$

$\Rightarrow \frac{2 a^{2} - a + 1}{5} \to N a t u r a l n u m b e r$

Let 2a² – a + 1 = 5x

D = 1 – 4 (2) (1 – 5n)

= 40n – 7, which is not $4 λ o r 4 λ + 1 f r o m .$

As each square form is $4 λ o r 4 λ + 1$

<math> <mrow> <mi>M</mi> <mi>e</mi> <mi>a</mi> <mi>n</mi> <mo>=</mo> <mfrac> <mrow> <mn>3</mn> <mo>+</mo> <mn>1</mn> <mn>2</mn> <mo>+</mo> <mn>7</mn> <mo>+</mo> <mi>a</mi> <mo>+</mo> <mrow> <mo> (</mo> <mrow> <mn>4</mn> <mn>3</mn> <mo>−</mo> <mi>a</mi> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>5</mn> </mrow> </mfrac> <mo>=</mo> <mn>1</mn> <mn>3</mn> </mrow> </math>  Variance =  <math> <mrow> <mfrac> <mrow> <msup> <mrow> <mn>3</mn> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>+</mo> <mn>1</mn> <msup> <mrow> <mn>2</mn> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>+</mo> <msup> <mrow> <mn>7</mn> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>+</mo> <msup> <mrow> <mi>a</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>+</mo> <msup> <mrow> <mrow> <mo> (</mo> <mrow> <mn>4</mn> <mn>3</mn> <mo>−</mo> <mi>a</mi> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> <mrow> <mn>5</mn> </mrow> </mfrac> <mo>−</mo> <msup> <mrow> <mrow> <mo> (</mo> <mrow> <mn>1</mn> <mn>3</mn> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </math>   <math> <mrow> <mo>⇒</mo> <mfrac> <mrow> <mn>2</mn> <msup> <mrow> <mi>a</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>−</mo> <mi>a</mi> <mo>+</mo> <mn>1</mn> </mrow> <mrow> <mn>5</mn> </mrow> </mfrac> <mo>→</mo> <mi>N</mi> <mi>a</mi> <mi>t</mi> <mi>u</mi> <mi>r</mi> <mi>a</mi> <mi>l</mi> <mtext> </mtext> <mtext> </mtext> <mi>n</mi> <mi>u</mi> <mi>m</mi> <mi>b</mi> <mi>e</mi> <mi>r</mi> </mrow> </math>  Let 2a2 – a + 1 = 5xD = 1 – 4 (2) (1 – 5n)= 40n – 7, which is not  <math> <mrow> <mn>4</mn> <mi>λ</mi> <mtext> </mtext> <mtext> </mtext> <mi>o</mi> <mi>r</mi> <mtext> </mtext> <mtext> </mtext> <mn>4</mn> <mi>λ</mi> <mo>+</mo> <mn>1</mn> <mtext> </mtext> <mtext> </mtext> <mi>f</mi> <mi>r</mi> <mi>o</mi> <mi>m</mi> <mo>.</mo> </mrow> </math>  As each square form is  <math> <mrow> <mn>4</mn> <mi>λ</mi> <mtext> </mtext> <mtext> </mtext> <mi>o</mi> <mi>r</mi> <mtext> </mtext> <mtext> </mtext> <mn>4</mn> <mi>λ</mi> <mo>+</mo> <mn>1</mn> </mrow> </math>

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The number of values of $a \in N$ such that the variance of 3, 7, 12, a, 43 -a is a natural number is:

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