Question

1. The following bell curves belong to probability distributions of three dif- ferent phenomena in nature. Which curve has the largest standard deviation o? 2. The article "Reliability of Domestic-Waste Biofilm Reactors" (J. of Envir. Engr., 1995: 785-790) suggests that substrate concentration of influent to a reactor is normally distributed with = 0.3 and 0 = 0.06. a. What is the probability that the concentration exceeds 25? b. What is the probability that the concentration is at most .10?

Answer 1

Therebor, The larger green colour standred carre bell deviation hos & probability 0.25 is that the concentration 0.7968 exceeds & probability most that 0.10 the is concentration 0.0004 is at

let we have given bell curves belong to Probability distrubutions of three diffrent phenomena in natue. there are three types of knetosis 6 lepto kustic & plattykuetic ③ meso kuɛtic. In lepto kuetic characteristics of this type of distrubution is one with long tails Couliers) the pretix of lepto -" means skinny" making the shape of leptokuetic dist? easier to remember. In platty Kuetic distrubutions have shost tails (paucity of outliers) the pretix of platy - means broad" for investors Platty kurtic return dist are stable and predictable in the sense e that there will rarely be extreme Coutliers) return. In mesokuetic distrubation has kuętosis statistic similar to that of the normal distrubution

In our problem, green colour bell Turre is reptoku etic Red Colour bell tume is meso kartic Black Colour bell curre is Platty karetic Here green colour bell cume is one of the long tail contliers) Red Colour bell curre is similar to the normal dist and Black Colour bell curre is more peaked thon the normal distrubution. there sore, it conclude that, green colour bell curre which has larger standred deviation (6) than the others. green colour bell curre has larger standred deviation (6) Here we have given the aeticle "Reliability of domestic waste Biofilm Reactors" suggests that substrate concentration of in Huert to a reactor is normally distrubuted with = 0.3 & 6 0.06

Here we have to probability that the find concentration exceeds 0:25 ie p(x > 0.25) PC x > 0.25) = P(x-ll 0.250.3 0106 " = P – 0.833) = PI - P(Z < 0.83 ) p( x 70125) - - 0.2032 = 0.7968 also we have to obtain. probability that the concentration is at most 0.10 - ie PEx 40.10) PCXL 040) - P(x-l colo-0.3) 0.06 2. P(Z <- 31383) normal probability table By wing p( x < 0.10) - 0.0004