Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

The mean and standard deviation of $40$ observations are $30$ and $5$ respectively. It was noticed that two of these observations $12$ and $10$ were wrongly recorded. If $\sigma$ is the standard deviation of the data after omitting the two wrong observations from the data, then $38 \sigma^{2}$ is equal to____________.
Option: 1
$238$

Option: 2
-

Option: 3
-

Option: 4
-

Answers (1)

best_answer

$\mathrm{\text { Mean }=30} \\$

$\mathrm{\Rightarrow \frac{\sum x}{40}=30 \Rightarrow \sum x=1200} \\$

$\mathrm{\text { Variance }=25} \\$

$\mathrm{\Rightarrow \frac{\sum x^{2}}{40}-30^{2}=25} \\$

$\mathrm{\Rightarrow \sum x^{2}=37000 }\\$

$\mathrm{\text { Correct } \sum x=1200-12-10=1178} \\$

$\mathrm{\text { Correct } \sum x^{2}=37000-144-100=36756 }$

$\therefore 38 \sigma^{2} =38\left[\frac{36756}{38}-\left(\frac{1178}{38}\right)^{2}\right] \\$

$=238$

Hence answer is 238

Posted by

rishi.raj

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

50 mL of 0.2 M ammonia solution is treated with 25 mL of 0.2 M HCl. If pK_b of ammonia solution is 4.75, the pH of the mixture will be :
Option: 1 3.75
Option: 2 4.75

Trending Articles/News

anam-khan

JEE Main 2025: Most important chapters for physics, chemistry, mathematics to score good marks

anam-khan

JEE Main 2025 session 2 dates clash with ISC, PSEB 12th exams; no impact on major science subjects

anam-khan

Delhi police EOW books FIITJEE for fraud; 192 complaints filed

Latest Question