Statistics And Probability Question 354
Question: The mean and S.D. of the marks of 200 candidates were found to be 40 and 15 respectively. Later, it was discovered that a score of 40 was wrongly read as 50. The correct mean and S.D. respectively are
Options:
A) $ 14.98,39.95 $
B) $ 39.95,14.98 $
C) $ 39.95,224.5 $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Corrected $ \Sigma x=40\times 200-50+40=7990 $
$ \therefore $ Corrected $ \bar{x}=7990/200=39.95 $ Incorrect $ \Sigma x^{2}=n[{{\sigma }^{2}}+{{\bar{x}}^{2}}]=200[15^{2}+40^{2}] $ $ =365000 $ Corrected $ \Sigma x^{2}=365000-2500+1600=364100 $
$ \therefore $ Corrected $ \sigma =\sqrt{\frac{364100}{200}-{{(39.95)}^{2}}} $ $ =\sqrt{(1820.5-1596)}=\sqrt{224.5}=14.98. $
BETA
