Let X be the score on a test. X follows normal distribution with mean 63.
The third quartile q3 = 80
The third quartile is the value that separates data so that 3/4 = 0.75 area lies below q3 and 1/4 =0.25 area lies above q3
Now we know that q3 =80.
P(x < 80) = 0.75
Using z score table we will find z value corresponding to probability 0.75
As there is nor exact probability value 0.75 in table we will consider most closest value which is 0.7486. The z score corresponding to 0.7486 is 0.67
Now we will use z=0.67, mean =63 and x=80 to find standard deviation
[tex] z = \frac{x - mean}{standard deviation} [/tex]
[tex] 0.67 = \frac{80- 63}{standard deviation} [/tex]
0.67 = 17 / standard deviation
standard deviation = 17/0.67 = 25.37
The population standard deviation is 25.37