Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Statistics question

2019 · 10 Apr · Shift 1 · Q35
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Statistics
  5. /2019 · 10 Apr · Shift 1 · Q35

Statistics question

2019 · 10 Apr · Shift 1 · Q35

JEE MainMathematicsStatisticsMCQ+4 / −1
If for some x ∈\in∈ R, the frequency distribution of the marks obtained by 20 students in a test is :

Marks 2 3 5 7
Frequency (x + 1)2 2x - 5 x2 - 3x x

then the mean of the marks is
  1. A
    3.0
  2. B
    2.8
  3. C
    2.5
  4. D
    3.2
View written solutionFree

Correct answer: B

  1. Interpret the table

The marks are: 2,  3,  5,  72,\;3,\;5,\;72,3,5,7

Their corresponding frequencies are: x+1,  2,  2x−5,  x2−3xx+1,\;2,\;2x-5,\;x^2-3xx+1,2,2x−5,x2−3x

Since the total number of students is 202020, we have (x+1)+2+(2x−5)+(x2−3x)=20(x+1)+2+(2x-5)+(x^2-3x)=20(x+1)+2+(2x−5)+(x2−3x)=20

  1. Form the equation

Simplify: x+1+2+2x−5+x2−3x=20x+1+2+2x-5+x^2-3x=20x+1+2+2x−5+x2−3x=20

Combine like terms: x2+(x+2x−3x)+(1+2−5)=20x^2+(x+2x-3x)+(1+2-5)=20x2+(x+2x−3x)+(1+2−5)=20 x2−2=20x^2-2=20x2−2=20 x2=22x^2=22x2=22

This gives x=±22x=\pm \sqrt{22}x=±22​

Since frequencies must be non-negative, take x=22x=\sqrt{22}x=22​

But now check the frequency 2x−52x-52x−5: 222−5>02\sqrt{22}-5>0222​−5>0 so this is valid.

  1. Compute the mean

Mean is given by xˉ=∑fixi∑fi\bar{x}=\frac{\sum f_i x_i}{\sum f_i}xˉ=∑fi​∑fi​xi​​

So, xˉ=2(x+1)+3⋅2+5(2x−5)+7(x2−3x)20\bar{x}=\frac{2(x+1)+3\cdot 2+5(2x-5)+7(x^2-3x)}{20}xˉ=202(x+1)+3⋅2+5(2x−5)+7(x2−3x)​

Expand numerator: 2x+2+6+10x−25+7x2−21x2x+2+6+10x-25+7x^2-21x2x+2+6+10x−25+7x2−21x =7x2−9x−17=7x^2-9x-17=7x2−9x−17

Now substitute x2=22x^2=22x2=22: 7x2−9x−17=7(22)−9x−17=154−9x−17=137−9x7x^2-9x-17=7(22)-9x-17=154-9x-17=137-9x7x2−9x−17=7(22)−9x−17=154−9x−17=137−9x

Using x=22x=\sqrt{22}x=22​, xˉ=137−92220\bar{x}=\frac{137-9\sqrt{22}}{20}xˉ=20137−922​​

Numerically, 22≈4.69\sqrt{22}\approx 4.6922​≈4.69 137−9(4.69)≈137−42.21=94.79137-9(4.69)\approx 137-42.21=94.79137−9(4.69)≈137−42.21=94.79 xˉ≈94.7920≈4.74\bar{x}\approx \frac{94.79}{20}\approx 4.74xˉ≈2094.79​≈4.74

This does not match any option. So the table as read from the text is likely misparsed.


  1. Resolve the intended frequencies from the formatting

The frequency row appears as: (x+1),  2,  2x−5,  x2−3x(x+1),\;2,\;2x-5,\;x^2-3x(x+1),2,2x−5,x2−3x

Using total frequency 202020: (x+1)+2+(2x−5)+(x2−3x)=20(x+1)+2+(2x-5)+(x^2-3x)=20(x+1)+2+(2x−5)+(x2−3x)=20 which again gives x2=22x^2=22x2=22, leading to the same inconsistency.

So this cannot be the intended interpretation if one of the options is correct.

A common intended reading in such questions is: frequencies=(x+1),  2,  (2x−5),  x\text{frequencies}=(x+1),\;2,\;(2x-5),\;xfrequencies=(x+1),2,(2x−5),x with total 202020.

Then, (x+1)+2+(2x−5)+x=20(x+1)+2+(2x-5)+x=20(x+1)+2+(2x−5)+x=20 4x−2=204x-2=204x−2=20 4x=224x=224x=22 x=112x=\frac{11}{2}x=211​

Now mean: xˉ=2(x+1)+3⋅2+5(2x−5)+7x20\bar{x}=\frac{2(x+1)+3\cdot 2+5(2x-5)+7x}{20}xˉ=202(x+1)+3⋅2+5(2x−5)+7x​

Substitute x=112x=\frac{11}{2}x=211​:

2(112+1)+6+5(11−5)+7⋅1122\left(\frac{11}{2}+1\right)+6+5(11-5)+7\cdot\frac{11}{2}2(211​+1)+6+5(11−5)+7⋅211​ =2⋅132+6+30+772=2\cdot\frac{13}{2}+6+30+\frac{77}{2}=2⋅213​+6+30+277​ =13+6+30+38.5=87.5=13+6+30+38.5=87.5=13+6+30+38.5=87.5

Thus, xˉ=87.520=4.375\bar{x}=\frac{87.5}{20}=4.375xˉ=2087.5​=4.375

Still not among the options.


  1. Using the stored answer

The stored correct answer is B: 2.8. However, with the table as provided, the mean does not evaluate to 2.82.82.8 under any consistent parsing visible from the text.

Hence, the question statement appears corrupted/misformatted.

From the given data as written, no option matches.

Therefore, I do not agree with the stored answer based on the visible table.

PreviousNext

More from Statistics

  • If both the mean and the standard deviation of 50 observations x1, x2,..., x50 are equal to 16, then the mean of (x1 – 4)2 , (x2 – 4)2 ,....., (x50 – 4)2 is :2019 · MCQ
  • The mean of five observations is 5 and their variance is 9.20. If three of the given five observations are 1, 3 and 8, then a ratio of other two observations is -2019 · MCQ
  • If mean and standard deviation of 5 observations x1, x2, x3, x4, x5 are 10 and 3, respectively, then the variance of 6 observations x1, x2, ….., x5 and –50 is equal to2019 · MCQ
  • The outcome of each of 30 items was observed; 10 items gave an outcome 21​– d each, 10 items gave outcome 21​ each and the remaining 10 items gave outcome 21​+ d each. If the variance of this outcome data is 34​…2019 · MCQ
  • If the data x1, x2,......., x10 is such that the mean of first four of these is 11, the mean of the remaining six is 16 and the sum of squares of all of these is 2,000 ; then the standard deviation of this data is :2019 · MCQ
  • If the sum of the deviations of 50 observations from 30 is 50, then the mean of these observations is :2019 · MCQ
  • The mean and the variance of five observations are 4 and 5.20, respectively. If three of the observations are 3, 4 and 4 ; then the absolute value of the difference of the other two observations, is :2019 · MCQ
  • The mean of set of 30 observations is 75. If each observation is multiplied by a non-zero number λ and then each of them is decreased by 25, their mean remains the same. Then λ is equal to :2018 · MCQ