Question 1026355
The mean would just be {{{(.74+.98)/2 = 0.86}}}.

The standard deviation would be {{{sqrt((1/12)(.98 - .74)^2) = sqrt(.0048) = .0693}}}.

The probability that the chlorine concentration will exceed 0.80 ppm is {{{(.98 - .80)*(1/(.98-.74)) = 0.18/0.24 = 0.75}}}.

The probability that the chlorine concentration will be between 0.80 ppm and 0.90 ppm is {{{(.90 - .80)*(1/(.98-.74)) = 0.10/0.24 = 0.417}}}.