Answer: 0.8956
Explanation:
Given that data is normally distributed :
Mean concentration (m) = 9.1 picomoles / litre
Standard deviation (σ) = 3.5 picomoles / litre
Probability that concentration is less than 13.5 picomoles per Litre = X
X < 13.5
Usingvthe Z relation :
Zscore = (x - m) / σ
Zscore = (13.5 - 9.1) / 3.5
Zscore = 4.4 / 3.5
Zscore = 1.257
P(Z < 1.257) = 0.8956