Skip to main content
Back to STATISTICS

Your Progress

0 / 141 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • MATHEMATICS • STATISTICS

Statistics practice · Medium

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

Take a chapter quiz →
Showing 91-120 of 141 questions

Preparing practice and saved progress…

91 Medium
91 of 141
Analyzing +4 | -1
If r1r_1 and r2r_2 are regression coefficients of yy on xx and xx on yy respectively, then
View answer and explanation
Correct option: C
r1+r2≥2rr_1+r_2 \geq 2 r
Option C (r1+r2≥2rr_1+r_2 \geq 2 r) is correct because the arithmetic mean of the regression coefficients is greater than or equal to the correlation coefficient.
92 Medium
92 of 141
Application +4 | -1
Let 9=x1<x2<…<x79=x_{1} < x_{2} < \ldots < x_{7} be in an A.P. with common difference dd. If the standard deviation of x1,x2…x7x_{1}, x_{2} \ldots x_{7} is 4 and the mean is xˉ\bar{x}, then xˉ+x6\bar{x}+x_{6} is equal to:
View answer and explanation
Correct option: B
34
Option B (34) is correct because it is the sum of the mean and the 6th term of the A.P., calculated using the given standard deviation.
93 Medium
93 of 141
Application +4 | -1
If a variate takes values a,ar,ar2,…,arn−1a, a r, a r^2, \ldots, a r^{n-1}, then which of the relation between means bold?
View answer and explanation
Correct option: A
AH=G2A H=G^2
Option A (AH=G2A H=G^2) is correct because the Arithmetic mean (A), Geometric mean (G), and Harmonic mean (H) of a geometric progression satisfy this relationship.
94 Medium
94 of 141
Application +4 | -1
The mean and variance of 5 observations are 5 and 8 respectively. If 3 observations are 1,3,51,3,5, then the sum of cubes of the remaining two observations is
View answer and explanation
Correct option: D
1072
The correct answer is D (1072) because the sum of cubes of the remaining two observations is calculated using the given mean and variance.
95 Medium
95 of 141
Application +4 | -1
The median of 100 observations grouped in classes of equal width is 25 . If the median class interval is 20−3020-30 and the number of observations less than 20 is 45 , then the frequency of median class is
View answer and explanation
Correct option: A
10
Option A (10) is correct. The frequency of the median class is calculated using the median formula for grouped data.
96 Medium
96 of 141
Application +4 | -1
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
View answer and explanation
Correct option: C
80%80\%
Option C (80%80\%) is correct because it is the percentage of boys calculated using the weighted average of the marks.
97 Medium
97 of 141
Application +4 | -1
The quartile deviation of daily wages (in ₹) of 11 persons given below, is
140,145,130,165,160,125,150,170,175,120,180140,145,130,165,160,125,150,170,175,120,180
View answer and explanation
Correct option: C
20
Option C (20) is correct because the quartile deviation is calculated as half the difference between the third and first quartiles.
98 Medium
98 of 141
Application +4 | -1
Given the following frequency distribution with some missing frequencies.
ClassFrequency10−2018020−30−−30−403440−5018050−6013660−70−−70−8050\begin{array}{cccccccccccccccccccc} {{\text{Class}}}&{{\text{Frequency}}} \\ {10 - 20}&{180} \\ {20 - 30}&{ - - } \\ {30 - 40}&{34} \\ {40 - 50}&{180} \\ {50 - 60}&{136} \\ {60 - 70}&{ - - } \\ {70 - 80}&{50} \end{array}

If the total frequency is 685 and median is 42.6 then the missing frequencies are
View answer and explanation
Correct option: C
82, 23
Option C (82, 23) is correct because solving the equations for the total frequency and median yields these missing frequencies.
99 Medium
99 of 141
Application +4 | -1
The mean and s.d. of a sample of size 10 were found
to be 9.5 and 2.5 respectively. Later on, an additional
observation 15 was added to the original data. The
s.d. of 11 observation is
View answer and explanation
Correct option: C
2.86
Option C (2.86) is correct because it is the standard deviation calculated after adding the additional observation and correcting the mean and variance.
100 Medium
100 of 141
Application +4 | -1
If mean and variance of observations 60,60,44,58,68,56,α,β60,60,44,58,68,56, \alpha, \beta are 58 and 66.2 respectively then α2+β2\alpha^2+\beta^2 is equal to
View answer and explanation
Correct option: B
7181.6
The correct answer is B; 7181.6. The problem involves using the formulas for mean and variance to find the values of α\alpha and β\beta, and then calculating α2+β2\alpha^2 + \beta^2.
101 Medium
101 of 141
Analyzing +4 | -1
Let r be the range and S2=1n−1∑i=1n(xi−xˉ)2{S^2} = \frac{1}{{n - 1}}\sum \limits_{i = 1}^n {\left( {{x_i} - \bar x} \right)^2} be the S.D. of a set observations x1,x2,…,xn,{x_1},{x_2}, \ldots ,{x_n}, then
View answer and explanation
Correct option: A
S≤rnn−1S \le r\sqrt {\frac{n}{{n - 1}}}
Option A (S≤rnn−1S \le r\sqrt {\frac{n}{{n - 1}}}) is correct because it represents the relationship between the standard deviation (S) and the range (r) of a set of observations.
102 Medium
102 of 141
Analyzing +4 | -1
Let the mean and variance of 12 observations be 92\frac{9}{2} and 4 respectively. Later on, it was observed that two observations were considered as 9 and 10 instead of 7 and 14 respectively. If the correct variance is mn\frac{m}{n}, where mm and nn are coprime, then m+nm+n are coprime, then m+nm+n is equal to
View answer and explanation
Correct option: D
317
Option D, 317, is the correct value of m+nm+n where the corrected variance is mn\frac{m}{n} and mm and nn are coprime. This is calculated by first finding the correct mean and then the correct variance after adjusting for the incorrect observations.
103 Medium
103 of 141
Application +4 | -1
The mean and variance of a set of 15 numbers are 12 and 14 respectively. The mean and variance of another set of 15 numbers are 14 and σ2\sigma^{2} respectively. If the variance of all the 30 numbers in the two sets is 13 , then σ2\sigma^{2} is equal to
View answer and explanation
Correct option: A
10
Option A (10) is correct because it is the variance of the second set of numbers, calculated using the combined variance formula.
104 Medium
104 of 141
Application +4 | -1
If the mean and the standard deviation of the data 3, 5, 7, aa, bb are 5 and 2 respectively, then aa and bb are the roots of the equation
View answer and explanation
Correct option: C
x2−10x+19=0{x^2} - 10x + 19 = 0
Option C (x2−10x+19=0{x^2} - 10x + 19 = 0) is correct because the equation is formed using the calculated sum and product of the roots aa and bb from the given mean and standard deviation.
105 Medium
105 of 141
Application +4 | -1
The mean and standard deviation of a sample of size 20 are found 10 and 2 respectively. If one item 8 is replaced by 12 then
View answer and explanation
Correct option: B
Standard deviation decreases
Option B (Standard deviation decreases) is correct because replacing 8 with 12 increases the mean by a small amount, but the overall spread of the data decreases, leading to a lower standard deviation.
106 Medium
106 of 141
Application +4 | -1
If the average of the numbers 148, 146, 144, 142, ... in AP, be 125, then the total numbers in the series will be
View answer and explanation
Correct option: B
24
Option B (24) is correct because the total number of terms in the AP series is calculated using the average and the properties of arithmetic progressions.
107 Medium
107 of 141
Understanding +4 | -1
The mean and variance for first n{n} natural numbers are respectively
View answer and explanation
Correct option: A
Mean =n+12=\frac{n+1}{2}, variance =n2−112=\frac{n^2-1}{12}
Option A (Mean =n+12\frac{{n+1}}{{2}}, variance =n2−112\frac{{n^2-1}}{{12}}) is correct because it uses the standard formulas for the mean and variance of the first n natural numbers.
108 Medium
108 of 141
Application +4 | -1
The mean square deviation of nn observations x1,x2,…,xnx_1, x_2, \ldots, x_n about -2 and 2 are 18 and 10 respectively. Then S.D. of the given set is
View answer and explanation
Correct option: C
3
Option C (3) is correct because the standard deviation is calculated correctly using the given mean square deviations.
109 Medium
109 of 141
Application +4 | -1
The mean and variance of 7 observations are 8 and 16 respectively. If the observation are 2,4,10,12,14,x,y2,4,10,12,14, x, y, then value of xyx y is
View answer and explanation
Correct option: D
48
Option D (48) is correct because it is the product of the two unknown observations (x and y) calculated using the given mean and variance of the 7 observations.
110 Medium
110 of 141
Understanding +4 | -1
If xˉ1\bar{x}_1 and σ1\sigma_1 are the mean and standard deviation of the first distribution and xˉ2\bar{x}_2 and σ2\sigma_2 are the mean and standard deviation of the second distribution.

I. CV (1st distribution)=σ1x‾1×100=\frac{\sigma_1}{\overline{{x}}_1} \times 100

II. CV (2nd distribution)=x‾2σ2×100=\frac{\overline{{x}}_2}{\sigma_2} \times 100

III. For x‾1=x‾2\overline{{x}}_1=\overline{{x}}_2, the series with lesser value of standard deviation is said to be more variable than the other.

IV. For x‾1=x‾2\overline{{x}}_1=\overline{{x}}_2, the series with greater value of standard deviation is said to be more consistent than the other.
View answer and explanation
Correct option: A
Only I is true
Option A (Only I is true) is correct because the coefficient of variation (CV) for the first distribution is indeed calculated as the standard deviation divided by the mean, multiplied by 100.
111 Medium
111 of 141
Application +4 | -1
The AM of n observations is M. If the sum of n-4 observations is a, then the mean of remaining 4 observations is
View answer and explanation
Correct option: A
n  M−a4\frac{{n\;M - a}}{4}
Option A (n;M−a4\frac{{n\\;M - a}}{4}) is correct because it represents the mean of the remaining 4 observations calculated by subtracting the sum of n-4 observations from the total sum and dividing by 4.
112 Medium
112 of 141
Application +4 | -1
The sum of 100 observations and the sum of their squares are 400 and 2475
respectively. Later on, three observations 3, 4 and 5 were found to be incorrect. If the
incorrect observations are omitted, then the variance of the remaining observations is
View answer and explanation
Correct option: C
9
Option C (9) is correct because it represents the variance of the remaining observations after correcting the sum and sum of squares by removing the incorrect observations.
113 Medium
113 of 141
Application +4 | -1
For a group containing 100 observation, the mean and
standard deviation are 8 and 10.5.\sqrt{10.5}. For 50 observations selected from these 100 observations, the mean
and standard deviation are 10 and 2 respectively . The
standard deviation of the other half is
View answer and explanation
Correct option: B
3
Option B (3) is correct because, using the combined variance formula and the given information, we can solve for the standard deviation of the other half.
114 Medium
114 of 141
Application +4 | -1
If the variance of the frequency distribution is 160 , then the value of c∈Nc \in N is
xc2c3c4c5c6cf211111\boxed{\begin{array}{cccccccccccccccccccc} x&c&{2c}&{3c}&{4c}&{5c}&{6c} \\ f&2&1&1&1&1&1 \end{array}}
View answer and explanation
Correct option: C
7
Option C (7) is correct because it is the value of c that makes the variance of the given frequency distribution equal to 160.
115 Medium
115 of 141
Understanding +4 | -1
A uniform rod of weight WW and length 2l2 l is resting in a smooth spherical bowl of radius rr. The rod is inclined to the horizontal at an angle of
View answer and explanation
Correct option: A
0∘0^{\circ}
Option A (0∘0^{\circ}) is correct because with a frictionless surface, the rod will rest at the minimum inclination, which is horizontal.
116 Medium
116 of 141
Application +4 | -1
If the mean deviation of number 1,1+d,1+2d,…..,1+100d1,1 + d,1 + 2d, \ldots ..,1 + 100d from their mean is 255, then the d is equal to
View answer and explanation
Correct option: C
10.1
Option C (10.1) is correct because the value of d is calculated using the formula for mean deviation and the given mean deviation value.
117 Medium
117 of 141
Understanding +4 | -1
Consider any set of observations x1,x2,x3,…,x101x_1, x_2, x_3, \ldots, x_{101}; it being given that x1<x2<x3<…<x100<x101x_1 < x_2 < x_3 < \ldots < x_{100} < x_{101}, then the mean deviation of this set of observations about a point kk is minimum, when kk equals
View answer and explanation
Correct option: B
x51x_{51}
Option B (x51x_{51}) is correct because the mean deviation is minimized when calculated about the median, which is x51x_{51} in this ordered set.
118 Medium
118 of 141
Application +4 | -1
If xˉ{\bar x} is the arithmetic mean of n independent variates x1,x2,x3,…,xn{x_1},{x_2},{x_3}, \ldots ,{x_n} each of the standard derivation σ, then variance (xˉ)\left( {\bar x} \right) is
View answer and explanation
Correct option: A
σ2n\frac{{{\sigma ^2}}}{n}
Option A (σ2n\frac{{\sigma ^2}}{n}) is correct because the variance of the mean of n independent variates is the variance of the individual variates divided by n.
119 Medium
119 of 141
Understanding +4 | -1
If xˉ1\bar{x}_1 and xˉ2\bar{x}_2 are means of two distribution such that xˉ1<xˉ2\bar{x}_1 < \bar{x}_2 and xˉ\bar{x} is the mean of the joint distribution, then
View answer and explanation
Correct option: D
xˉ1<xˉ<xˉ2\bar{x}_1 < \bar{x} < \bar{x}_2
Option D (xˉ1<xˉ<xˉ2\bar{x}_1 < \bar{x} < \bar{x}_2) is correct because the mean of the joint distribution lies between the means of the individual distributions when xˉ1<xˉ2\bar{x}_1 < \bar{x}_2.
120 Medium
120 of 141
Application +4 | -1
The greatest and least magnitude of the resultant of two forces of constant magnitude are FF and GG. When the forces act an angle 2α2 \alpha, the resultant in magnitudes is equal to
View answer and explanation
Correct option: A
F2cos⁡2α+G2sin⁡2α\sqrt{F^2 \cos ^2 \alpha+G^2 \sin ^2 \alpha}
Option A, F2cos⁡2α+G2sin⁡2α\sqrt{F^2 \cos ^2 \alpha+G^2 \sin ^2 \alpha}, is correct because it represents the magnitude of the resultant when the forces act at an angle 2α2 \alpha, given the greatest and least magnitudes F and G.
Revision notes and key takeaways

Review STATISTICS in your textbook or class notes, then use this set to check your understanding.

Showing 91-120 of 141 questions

Your practice

0 / 141 answered · 141 remaining · 0% accuracy

Progress is stored on this device. Sign in to save to your account.

✓ Correct · ✗ Incorrect · ○ Unanswered

Answer feedback and difficulty

Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.