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Question

Read the following frequency distribution for two series of observations and answer the two questions below.

Q. What is the difference between the mean of the frequency distribution of series I and II?

अवलोकनों की दो श्रेणीयों के लिए निम्नलिखित आवृत्ति वितरण पढ़ें और नीचे दिए गए दो प्रश्नों के उत्तर दें।

Q. उपरोक्त गद्यांश के संदर्भ में, निम्नलिखित में से कौन-सा/से कथन सही है/हैं?

श्रेणी I और II की आवृत्ति वितरण के बीच कितना अंतर है?

A

3.6
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B

0.7
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C

1.4
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D

Cannot be determined
निर्धारित नहीं किया जा सकता है
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Solution

The correct option is A
3.6

First let's find out the values of x and y.

2x+y = 100 - (12+16+10) = 62 ---(1)

x+y = 100 - (6+24+19) = 51 ---(2)

Subtract (1) from (2), we get, x = 11 and y = 40

Mean of the frequency distribution of series I= (12×15)+(16×25)+(10×35)+(22×45)+(40×55)100=41.2

mean of the frequency distribution of series II=(6×15)+(24×25)+(19×35)+(40×45)+(11×55)100=37.6

Difference = 41.2 - 37.6 = 3.6


पहले x और y का मान निकालें।

2x+y = 100 - (12+16+10) = 62 ---(1)

x+y = 100 - (6+24+19) = 51 ---(2)

समीकरण (1) को समीकरण (2) में से घटाने पर,

x = 11 और y = 40

श्रेणी I की आवृत्ति वितरण का माध्य = (12×15)+(16×25)+(10×35)+(22×45)+(40×55)100=41.2

श्रेणी II की आवृत्ति वितरण का माध्य =(6×15)+(24×25)+(19×35)+(40×45)+(11×55)100=37.6

अंतर = 41.2 - 37.6 = 3.6


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