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The Mean of 200 Items Was 50. Later On, It Was Discovered that Two Items Were Misread as 92 and 8 Instead of 192 and 88. Find the Correct Mean.

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Question

The mean of 200 items was 50. Later on, it was discovered that two items were misread as 92 and 8 instead of 192 and 88.
Find the correct mean.

Sum
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Solution

Given that the mean of 200 items was 50.
Mean = `sum_x/n`

⇒ 50 = `sum_x/200`

⇒ x = 10,000

Incorrect value of  `sum x`= 10,000
correct value of 
`sumx` = 10,000 - ( 92 + 8 ) + ( 192 + 88 )
= 10,000 - 100 + 280
= 10,180

Correct mean
= `("correct value of" sumx")/n` 

= `10180/200`

= 50.9

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Chapter 19: Mean and Median (For Ungrouped Data Only) - Exercise 19 (A) [Page 239]

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Selina Concise Mathematics [English] Class 9 ICSE
Chapter 19 Mean and Median (For Ungrouped Data Only)
Exercise 19 (A) | Q 11 | Page 239
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