Answer
Mean:
$y ̅ =1.19~nmol~ATP/mg$
Standard deviation:
$y ̃=0.184~nmol~ATP/mg$
Work Step by Step
$y ̅ =\frac{∑y_i}{n}=\frac{1.45+1.19+1.05+1.07}{4}=\frac{4.76}{4}=1.19$
$s=\sqrt {\frac{∑(y_i-y ̅ )^2}{n-1}}=\sqrt {\frac{(1.45-1.19)^2+(1.19-1.19)^2+(1.05-1.19)^2+(1.07-1.19)^2}{4-1}}=0.184$
