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Statistics

JEE Mathematics · 96 questions · Page 2 of 10 · Click an option or "Show Solution" to reveal answer

Q11
In an experiment with 15 observations on xx, then following results were available: ∑x2=2830\sum {{x^2}} = 2830, ∑x=170\sum x = 170 One observation that was 20 was found to be wrong and was replaced by the correct value 30. Then the corrected variance is :
A 188.66
B 177.33
C 8.33
D 78.00
Correct Answer
Option D
Solution

Given that, N = 15,

∑x2=2830, ∑x=170\sum {x{}^2} = 2830,\,\sum x = 170

As, 20 was replaced by 30 then,

∑x=170−20+30=180\sum x = 170 - 20 + 30 = 180

and

∑x2=2830−400+900=3330\sum {{x^2}} = 2830 - 400 + 900 = 3330

So, the corrected variance

=∑x2N−(∑xN)2= {{\sum {{x^2}} } \over N} - {\left( {{{\sum x } \over N}} \right)^2}
=333015−(18015)2= {{3330} \over {15}} - {\left( {{{180} \over {15}}} \right)^2}
=222−122= 222 - {12^2}
=222−144= 222 - 144
=78= 78
Q12
Consider the following statements: (a) Mode can be computed from histogram (b) Median is not independent of change of scale (c) Variance is independent of change of origin and scale. Which of these is/are correct?
A only (a)
B only (b)
C only (a) and (b)
D (a), (b) and (c)
Correct Answer
Option C
Solution

The statements are analyzed as follows : (a) Mode can be computed from histogram : This is correct.

The mode is the value that appears most frequently in a data set.

A histogram provides a graphical representation of the frequency of each data value.

The data value corresponding to the highest bar in a histogram is the mode. (b) Median is not independent of change of scale : This is correct.

The median is the middle value in a data set when the values are arranged in ascending or descending order.

When you change the scale (e.g., by multiplying all data points by a constant), the median changes as well. (c) Variance is independent of change of origin and scale : This is not correct.

Variance, which measures the dispersion of a set of data points, is affected by changes in scale (it is not independent of scale).

If you multiply all the values in a dataset by a constant, the variance gets multiplied by the square of that constant.

However, it is true that variance is independent of the change of origin (it remains the same if you add or subtract a constant from all data points in the set).

Therefore, the correct answer is Option C : only (a) and (b) are correct.

Q13
In a series of 2n observations, half of them equal aa and remaining half equal –a–a. If the standard deviation of the observations is 2, then ∣a∣|a| equals
A 2
B 2\sqrt 2
C 1n{1 \over n}
D 2n{{\sqrt 2 } \over n}
Correct Answer
Option A
Solution

Mean

(A)=a−a2n=0\left( A \right) = {{a - a} \over {2n}} = 0

Given standard deviation (S.D) = 2

∴   \therefore\,\,\,
∑(x−A)22n=2\sqrt {{{\sum {{{\left( {x - A} \right)}^2}} } \over {2n}}} = 2
⇒   (a−0)2+(a−0)2+.....+(0−a)22n=2\Rightarrow \,\,\,\sqrt {{{{{\left( {a - 0} \right)}^2} + {{\left( {a - 0} \right)}^2} + ..... + {{\left( {0 - a} \right)}^2}} \over {2n}}} = 2
⇒   a2+a2........2n ×2n=2\Rightarrow \,\,\,\sqrt {{{{a^2} + {a^2}........2n\,\times} \over {2n}}} = 2
⇒   2n . a22n=2\Rightarrow \,\,\,\sqrt {{{2n\,.\,{a^2}} \over {2n}}} = 2
⇒   a2=2\Rightarrow \,\,\,\sqrt {{a^2}} = 2
⇒   ∣a∣=2\Rightarrow \,\,\,\left| a \right| = 2
Q14
Let x1, x2,...........,xn be n observations such that ∑xi2=400\sum {x_i^2} = 400 and ∑xi=80\sum {{x_i}} = 80. Then a possible value of n among the following is
A 18
B 15
C 12
D 9
Correct Answer
Option A
Solution

As we know,

σ2≥0{\sigma ^2} \ge 0
∴   \therefore\,\,\,
∑xi2n−(∑xin)2≥0{{\sum {x_i^2} } \over n} - {\left( {{{\sum {{x_i}} } \over n}} \right)^2} \ge 0
⇒   400n−6400n2≥0\Rightarrow \,\,\,{{400} \over n} - {{6400} \over {{n^2}}} \ge 0
⇒   n≥16\Rightarrow \,\,\,n \ge 16
∴   \therefore\,\,\,

Possible value of n according to the option is = 18

Q15
If in a frequency distribution, the mean and median are 21 and 22 respectively, then its mode is approximately :
A 20.5
B 22.0
C 24.0
D 25.5
Correct Answer
Option C
Solution

Given that, Mean = 21 and median = 22 We know, Mode + 2 Mean = 3 Median ∴\therefore Mode = 3 ×\times 22 −- 2 ×\times 21 = 66 −- 42 = 24

Q16
Suppose a population A has 100 observations 101, 102,........, 200, and another population B has 100 observations 151, 152,......., 250. If VA and VB represent the variances of the two populations, respectively, then VAVB{{{V_A}} \over {{V_B}}} is
A 1
B 94{9 \over 4}
C 49{4 \over 9}
D 23{2 \over 3}
Correct Answer
Option A
Solution

Series A = 101, 102 ............

200 Series B = 151, 152 ............

250 Here series B can be obtained if we change the origin of A by 50 units.

And we know the variance does not change by changing the origin.

So,

    \,\,\,\,
VA=VB{V_A} = {V_B}
⇒     VAVB=1\Rightarrow \,\,\,\,\,{{{V_A}} \over {{V_B}}} = 1
Q17
The average marks of boys in a class is 52 and that of girls is 42. The average marks of boys and girls combined is 50. The percentage of boys in the class is
A 80
B 60
C 40
D 20
Correct Answer
Option A
Solution

Let x and y are number of boys and girls in a class respectively. ∴\therefore 52x + 42y = 50 (x + y) ⇒\Rightarrow 52x + 42y = 50x + 50y ⇒\Rightarrow 2x = 8y ⇒\Rightarrow x = 4y ∴\therefore Total no. of students = x + y = 4y + y = 5y ∴\therefore Percentage of boys =

4y5y×100 %{{4y} \over {5y}} \times 100\,\%

= 80%

Q18
The mean of the numbers a, b, 8, 5, 10 is 6 and the variance is 6.80. Then which one of the following gives possible values of a and b?
A a = 0, b = 7
B a = 5, b = 2
C a = 1, b = 6
D a = 3, b = 4
Correct Answer
Option D
Solution

Given that, Mean of a, b, 8, 5, 10 = 6

∴   \therefore\,\,\,
a+b+8+5+105=6{{a + b + 8 + 5 + 10} \over 5} = 6
⇒   \Rightarrow \,\,\,

a + b + 23 = 30

⇒   \Rightarrow \,\,\,

a + b = 7 ....... (1) Variance

=∑(xi−A)2n=6.8= {{\sum {{{\left( {{x_i} - A} \right)}^2}} } \over n} = 6.8
⇒   (6−a)2+(6−b)2+(6−8)2+(6−5)2+(6−10)25=6.5\Rightarrow \,\,\,{{{{\left( {6 - a} \right)}^2} + {{\left( {6 - b} \right)}^2} + {{\left( {6 - 8} \right)}^2} + {{\left( {6 - 5} \right)}^2} + {{\left( {6 - 10} \right)}^2}} \over 5} = 6.5
⇒   (6−a)2+(6−b)2+4+1+16=34\Rightarrow \,\,\,{\left( {6 - a} \right)^2} + {\left( {6 - b} \right)^2} + 4 + 1 + 16 = 34
⇒(6−a)2+(6−b)2=13\Rightarrow {\left( {6 - a} \right)^2} + {\left( {6 - b} \right)^2} = 13
⇒   a2+b2=25\Rightarrow \,\,\,{a^2} + {b^2} = 25

By using (1) we get,

a2+(7−a)2=25{a^2} + \left( {7 - a} \right){}^2 = 25
⇒   a2−7a+12=0\Rightarrow \,\,\,a{}^2 - 7a + 12 = 0
∴   \therefore\,\,\,

a = 3, 4 then b = 4, 3.

Q19
Statement - 1 : The variance of first n even natural numbers is n2−14{{{n^2} - 1} \over 4} Statement - 2 : The sum of first n natural numbers is n(n+1)2{{n\left( {n + 1} \right)} \over 2} and the sum of squares of first n natural numbers is n(n+1)(2n+1)6{{n\left( {n + 1} \right)\left( {2n + 1} \right)} \over 6}
A Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1
B Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1
C Statement-1 is true, Statement-2 is false
D Statement-1 is false, Statement-2 is true
Correct Answer
Option D
Solution

Let first n even natural numbers = 2,4, 6, 8 ...... 2n ∴\therefore Sum of those num = 2 + 4 + 6 + ..... 2n = 2 (1 + 2 + ..... n) =

2.n(n+1)22.{{n\left( {n + 1} \right)} \over 2}

= n (n + 1)

∴   \therefore\,\,\,

Mean

(x‾)=n(n+1)n\left( {\overline x } \right) = {{n\left( {n + 1} \right)} \over n}
=n+1= n + 1
∴   \therefore\,\,\,

Variance

=1n∑xi2−(x‾)2= {1 \over n}\sum {x_i^2} - {\left( {\overline x } \right)^2}
=1n[22+42+......+(2n)2]−(n+1)2= {1 \over n}\left[ {{2^2} + {4^2} + ...... + {{\left( {2n} \right)}^2}} \right] - {\left( {n + 1} \right)^2}
=1n22[12+22+.......n2]−(n+1)2= {1 \over n}{2^2}\left[ {{1^2} + {2^2} + .......n{}^2} \right] - \left( {n + 1} \right){}^2
=4n[n(n+1)(2n+1)6]−(n+1)2= {4 \over n}\left[ {{{n\left( {n + 1} \right)\left( {2n + 1} \right)} \over 6}} \right] - {\left( {n + 1} \right)^2}
=(n+1)[2(2n+1)−3(n+1)]3= {{\left( {n + 1} \right)\left[ {2\left( {2n + 1} \right) - 3\left( {n + 1} \right)} \right]} \over 3}
=(n+1)(n−1)3= {{\left( {n + 1} \right)\left( {n - 1} \right)} \over 3}
=n2−13= {{{n^2} - 1} \over 3}

∴\therefore Statement 1 is false. Statement 2 is true as those are standard formula.

Q20
If the mean deviation of number 1, 1 + d, 1 + 2d,........, 1 + 100d from their mean is 255, then the d is equal to
A 20.0
B 10.1
C 20.2
D 10.0
Correct Answer
Option B
Solution

Mean

    (x‾)=Sum  of  numbersn\,\,\,\,\left( {\overline x } \right) = {{Sum\,\,of\,\,numbers} \over n}
=n2(a+l)n= {{{n \over 2}\left( {a + l} \right)} \over n}
=12(1+l+100d)= {1 \over 2}\left( {1 + l + 100d} \right)
=1+50 d.= 1 + 50\,d.

Mean deviation (M.D)

=1n∑i=1101∣xi−x‾∣= {1 \over n}\sum\limits_{i = 1}^{101} {\left| {{x_i} - \overline x } \right|}
=1101= {1 \over {101}}

[ 50d + 49d + 48d + .......d + 0 + ..... + 50d]

=1101.2d(1+2+....+50)= {1 \over {101}}.2d\left( {1 + 2 + .... + 50} \right)
=1101.2d.50×512= {1 \over {101}}.2d.{{50 \times 51} \over 2}
=50×51 d101= {{50 \times 51\,d} \over {101}}

Given that M.D = 255

∴   \therefore\,\,\,
50×51 d101=255{{50 \times 51\,d} \over {101}} = 255
⇒   d=101×25551×50\Rightarrow \,\,\,d = {{101 \times 255} \over {51 \times 50}}
=10.1= 10.1
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