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तामिळनाडू बोर्ड ऑफ सेकेंडरी एज्युकेशनएस.एस.एल.सी. (इंग्रजी माध्यम) इयत्ता १०

The frequency distribution is given below. x k 2k 3k 4k 5k 6k f 2 1 1 1 1 1 In the table, k is a positive integer, has a variance of 160. Determine the value of k. - Mathematics

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प्रश्न

The frequency distribution is given below.

x k 2k 3k 4k 5k 6k
f 2 1 1 1 1 1

In the table, k is a positive integer, has a variance of 160. Determine the value of k.

बेरीज

उत्तर

Assumed mean = 3k

x fi d = xi − A
= xi − 3k
fidi fidi2
k 2 − 2k − 4k 8k2
2k 1 − k − k k2
3k 1 0 0 0
4k 1 k k k2
5k 1 2k 2k 4k2
6k 1 3k 3k 9k2
  `sumf_"i"` = 7   `sumf_"i""d"_"i"` = k `sumf_"i""d"_"i"^2` = 23k2

Here `sumf_"i"` = 7, `sumf_"i""d"_"i"` = k, `sumf_"i""d"_"i"^2` = 23k2

Variance = 160

`(sumf_"i""d"_"i"^2)/(sumf_"i") - ((sumf_"i""d"_"i")/(sumf_"i"))^2` = 160

`(23"k"^2)/7 - ("k"/7)^2` = 160

`(23"k"^2)/7 - ("k"^2/49)` = 160

⇒ `(161"k"^2 - "k"^2)/49` = 160

`(160"k"^2)/49` = 160

⇒ 160k2 = 160 × 49

k2 = `(160 xx 49)/160`

k = `sqrt(49)`

⇒ k = 7

The value of k = 7

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Measures of Dispersion
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पाठ 8: Statistics and Probability - Unit Exercise – 8 [पृष्ठ ३३२]

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सामाचीर कलवी Mathematics [English] Class 10 SSLC TN Board
पाठ 8 Statistics and Probability
Unit Exercise – 8 | Q 3 | पृष्ठ ३३२
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