There are 60 students in a class- The following is the frequency distribution of the marks obtained by the students in a test0 - x-2 Whline1 - x INhline2 - x- 2 Wline3 -x-1- 2 Whline4 - 2x lWhline5 - x-1 livhline lend array Where x is positive integer- Then the variance of the marks isA- 1-26B- 1-28C- 1-34D- 1-24

Sum of frequencies nbsp; ∑ f_i=60 x 2+x+x^2+(x+1)^2+2x+x+1=60 ⇒ 2x^2+7x 60=0 ⇒ (x 4)(2x+15) = 0 ∴ nbsp;x=4 (As x 2 is frequency, so x 2≥ 0) Now, [ ...

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Standard XI

Statistics

Spearman's Rank Coefficient for Quantitative Data

There are 60 ...

Question

There are $60$ students in a class. The following is the frequency distribution of the marks obtained by the students in a test

$\begin{matrix} Marks & Frequency 0 & x - 2 1 & x 2 & x^{2} 3 & (x + 1)^{2} 4 & 2 x 5 & x + 1 \end{matrix}$

Where $x$ is positive integer. Then the variance of the marks is

1.26

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1.28

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1.34

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1.24

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Solution

The correct option is A $1.26$
Sum of frequencies
$\sum f_{i} = 60$
$x - 2 + x + x^{2} + (x + 1)^{2} + 2 x + x + 1 = 60$
After simplifying we will get
$x = 4$

$\begin{matrix} x_{i} & f_{i} & f_{i} x_{i} & f_{i} x_{i}^{2} 0 & 2 & 0 & 0 1 & 4 & 4 & 4 2 & 16 & 32 & 64 3 & 25 & 75 & 225 4 & 8 & 32 & 128 5 & 5 & 25 & 125 \sum f_{i} x_{i} = 168 & \sum f_{i} x_{i}^{2} = 546 \end{matrix}$

Mean $(¯ x) = \frac{\sum f_{i} x_{i}}{\sum f_{i}} = \frac{168}{60} = 2.8$

Variance $(σ^{2}) = \frac{1}{N} \sum f_{i} x_{i}^{2} - (Mean)^{2}$
$= \frac{546}{60} - (2.8)^{2} = 1.26$

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Spearman's Rank Coefficient for Quantitative Data

Standard XI Statistics

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There are 60 students in a class. The following is the frequency distribution of the marks obtained by the students in a test MarksFrequency0x−21x2x23(x+1)242x5x+1 Where x is positive integer. Then the variance of the marks is

The correct option is A 1.26Sum of frequencies ∑fi=60 x−2+x+x2+(x+1)2+2x+x+1=60 After simplifying we will get x=4 xififixifix2i0200144421632643257522548321285525125∑fixi=168∑fix2i=546 Mean (¯x)=∑fixi∑fi=16860=2.8 Variance (σ2)=1N∑fix2i−(Mean)2 =54660−(2.8)2=1.26