slader 3. The daily surface concentration of carbonyl sulfide on the Indian Ocean is normally distributed, with a mean of 9.1 picomoles per liter and a standard deviation of 3.5 picomoles per liter. Find the probability that on a randomly selected day, the surface concentration of carbonyl sulfide on the Indian Ocean is less than 13.5 picomoles per liter.

0.8956

Step-by-step 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

square root of 36/6 = 2.449489743

square root of 16/6 = 1.632993162

square root of 6 = 2.449489743

the answer is square root of 16/6 because the square root of 36/6 is the same answer as square root of 6 = 2.449489743