Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

# 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.

#### 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,

$\sigma^2 = \frac{\sum f_i \left( x_i - \bar{X} \right)^2}{N} = \frac{9448 . 848}{2600} = 3 . 63$

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#### APPEARS IN

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