The PEFR of a group of 11 year old girls follow a normal distrubution with mean 300 1/min and standard deviation 20 1/min -
High-Yield Explanation
1 SD includes → 68% of values
2 SD includes → 95% of values
3 SD includes → 99.7% of values
In this question
Mean PEFR = 300 L/min
SD = 2 L/min
Area around 1SD on either side of mean (x ± 1SD) will include 68% of values, i.e.
x± 1SD=300±20
So, 68% of girls have PEFR between 280 & 320 L/min.
Area around 2SD on either side of mean (x ± 2SD) will include 95% of values, i.e.
x±2SD=300±40
So, 95% of girls have PEFR between 260 & 340 L/min.
Area around 3SD on either side of mean (x ± 3SD) will include 99.7% of values, i.e. -
x ±3SD = 300 ± 60
So, 99.7% of girls will have PEFR between 240 & 360 L/min.