# What was the standard deviation of this problem:Tim scored 59 on a Year 7 numeracy test, which was 1 standard deviation below the mean. On the same test, Francis scored 77, which was 2 standard...

What was the standard deviation of this problem:

Tim scored 59 on a Year 7 numeracy test, which was 1 standard deviation below the mean. On the same test, Francis scored 77, which was 2 standard deviation above the mean.

lfryerda | High School Teacher | (Level 2) Educator

Posted on

Since 59 is one standard deviation below the mean and 77 is two standard deviations above the mean, we see that the standard deviation must be one third of the difference between 77 and 59, which is (77-59)/3=6.

The standard deviation is 6.

Sources:

neela | High School Teacher | (Level 3) Valedictorian

Posted on

Let m and s be the mean and standard deviation.

Then by data,

Tim's score : m-s = 59

Francis' score: m+2s = 77

Francis' score -Tim's score: m+2s-(m-s) =77-59

=> 3s = 18

=> 3s/3 = 18/3

=> s = 6.

Substitute s = 6 in the first equation: m-6 = 59

=> m = 59+6 = 65.

The standard deviation = 6. The mean = 65.