Question

In an experiment, the target bacteria can only live in the solution with 0.7-0.9ml/L NaCl. Your solution has mean 0.8ml/L but standard deviation 0.05ml/L NaCl. What is a lower bound of the probability that the bacteria will survive?

Answer 1

The probability of survival is at least 75%

Step-by-step explanation:

Since we do not know the probability distribution of the concentration of the solution, we can not determine the exact probability of survival. Nevertheless, we can use the Chebyshev inequality, that applies to any probability distribution, to calculate the lower bound.

It states that

P( | X - µ | ≤ kσ ) ≥ 1- 1/k²

this means that the probability that X is not further than k standard deviations from the mean (µ) is at least 1- 1/k² , where k >1

in our case,

k1= (X1 - µ) / σ = (0,9 - 0,8 ) / 0,05 = 2

k2=(X2 - µ) / σ = (0,7- 0,8 ) / 0,05 = -2

therefore k=k1=k2=2 and

P( | X - µ | ≤ 2σ ) ≥ 1- 1/2²= 3/4=75%

consequently, the lower bound of the probability of survival is 75%