Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11
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Table Below Shows the Frequency F With Which 'X' Alpha Particles Were Radiated from a Diskette X :0123456789101112f :5120338352553240827313943271042calculate the Mean and Variance. - Mathematics

Table below shows the frequency f with which 'x' alpha particles were radiated from a diskette 

x : 0 1 2 3 4 5 6 7 8 9 10 11 12
f : 51 203 383 525 532 408 273 139 43 27 10 4 2

Calculate the mean and variance.

 

 
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Solution

Mean,

\[\bar{X} = \frac{\sum f_i x_i}{\sum f_i} = \frac{10078}{2600} = 3 . 88\]
 

\[x_i\]
 

\[f_i\]
 
\[f_i x_i\]
 

\[x_i - \bar{X}\]
 

\[\left( x_i - \bar{X} \right)^2\]
 

\[f_i \left( x_i - \bar{X} \right)^2\]
0 51 0 −3.88 15.05 767.55
1 203 203 −2.88 8.29 1682.87
2 383 766 −1.88 3.53 1351.99
3 525 1575 −0.88 0.77 404.25
4 532 2128 0.12 0.014 7.448
5 408 2040 1.12 1.25 510
6 273 1638 2.12 4.49 1225.77
7 139 973 3.12 9.73 1352.47
8 43 344 4.12 16.97 729.71
9 27 243 5.12 26.21 707.67
10 10 100 6.12 37.45 374.5
11 4 44 7.12 50.69 202.76
12 2 24 8.12 65.93 131.86
 
 

\[\sum f_i = N = 2600\]
 
\[\sum f_i x_i = 10078\]
   
 

\[\sum f_i \left( x_i - \bar{X} \right)^2 = 9448 . 848\]

Variance,

\[\sigma^2 = \frac{\sum f_i \left( x_i - \bar{X} \right)^2}{N} = \frac{9448 . 848}{2600} = 3 . 63\]
 
  Is there an error in this question or solution?
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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 32 Statistics
Exercise 32.5 | Q 2 | Page 38
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