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NCERT Exemplar · Q11
Q.

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

Marks012345
Frequencyx−2x-2xxx2x^2(x+1)2(x+1)^22x2xx+1x+1

where xx is a positive integer. Determine the mean and standard deviation of the marks.

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The total frequency =60=60 gives x=4x=4; then the mean is 2.82.8 and the standard deviation is 1.26≈1.12\sqrt{1.26}\approx 1.12.

Step 1 — Find xx

The frequencies must add to 60:

(x−2)+x+x2+(x+1)2+2x+(x+1)=60.(x-2)+x+x^{2}+(x+1)^{2}+2x+(x+1)=60.

Expanding (x+1)2=x2+2x+1(x+1)^2=x^2+2x+1 and collecting terms:

2x2+7x=60  ⇒  2x2+7x−60=0.2x^{2}+7x=60\;\Rightarrow\;2x^{2}+7x-60=0.

x=−7±49+4804=−7±234.x=\frac{-7\pm\sqrt{49+480}}{4}=\frac{-7\pm 23}{4}.

The positive integer root is x=4x=4 (the other root is negative and rejected). So the frequencies are 2,4,16,25,8,52,4,16,25,8,5 (check: 2+4+16+25+8+5=602+4+16+25+8+5=60 ✓).

Step 2 — Mean

Marks mmfffmfmfm2fm^{2}
0200
1444
2163264
32575225
4832128
5525125
Total60168546

mˉ=∑fmN=16860=2.8.\bar{m}=\frac{\sum fm}{N}=\frac{168}{60}=2.8.

Step 3 — Standard deviation …

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