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How to solve this standard deviation?


Fishi

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  • Two different groups of test-takers received scores on the GXYZ standardized test. Group A''s scores had a normal distribution with a mean of 460 and a standard deviation of 20. Group B''s scores had a normal distribution with a mean of 520 and a standard deviation of 40. If each group has the same number of test-takers, what fraction of the test-takers who scored below 440 belonged to Group B?

a) 1/9

b) 1/8

c) 1/6

d) 4/17

e) 4/21

 

Can anybody tell me how to solve this?

 

Thanks!:tup:

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Is the answer 1/9 ?

 

Here is what I came up with:

 

Group A: Mean 460, SD =20 ,Example series: 400,420,440,460,480,500,520

Group A: Mean 520, SD =40 ,Example series: 400,440,480,520,560,600,640

 

 

Here I have taken example series of 7 numbers in each.. these series statisfy Mean and SD.

 

Now, lets get to the point.

 

only 400 is less than 440 in Series B. Total is 14. => 1/7 ( not given in answers).

 

If I take another no# on both sides for each series it becomes 2/18 =>1/9..

 

So answer can be any one of the following series : 1/7,1/9,3/22, 4/26...

 

1/9 suits in our case

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In a normally distributed data, about 68% of the values are within one standard deviation (SD) of the mean, about 95% of the values are within 2SD and about 99.7 % lie within 3SD.

 

Group A has mean of 460 and SD of 20, so one who scored less than 440 lies 1SD away from mean = (100-68)/2 = 16%

 

Group B has mean of 520 and SD of 40, so one who scored less than 440 lies 2SD away from mean = (100-95)/2 = 2.5%

 

Fraction of total test-takers from Group-B who scored less than 440

= 2.5/(2.5+16) = 2.5/18.5 can be approximated to 1/7.4 => answer: 1/8

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