Question

The ph of water samples from a specific lake is a random variable y with probability density function given by f (y) = $ (3/8)(7 − y) 2 , 5 ≤ y ≤ 7, 0, elsewhere. a find e(y ) and v(y ). b find an interval shorter than (5, 7) in which at least three-fourths of the ph measurements must lie. c would you expect to see a ph measurement below 5.5 very often? why?

Answer 1

Given that the ph of water samples from a specific lake is a random variable y with probability density function given by


`f(y)= \left \{ {{ (3)/(8)(7-y)^2 \ \ \ 5\leq y\leq7 } \atop {0 \ \ \ \ \ \ \ elsewhere}} \right.`

Part A:


`E(y)= \int\limits^\infty_(-\infty) {yf(y)} \, dy \\ \\ = \int\limits^7_5 {y \left((3)/(8)\right)(7-y)^2} \, dy\\ \\ = (3)/(8)\int\limits^7_5 (49y-14y^2+y^3)dy \\ \\ = (3)/(8) \left[ (49)/(2) y^2- (14)/(3) y^3+ (1)/(4) y^4\right]^7_5 \\ \\ = (3)/(8) [(1,200.5-1,600.67+600.25)-(612.5-583.33+156.25)] \\ \\ = (3)/(8) (200.08-185.42)= (3)/(8) (14.66)=5.5`



Part B:


`E(y^2)= \int\limits^\infty_(-\infty) {y^2f(y)} \, dy \\ \\ = \int\limits^7_5 {y^2 \left((3)/(8)\right)(7-y)^2} \, dy\\ \\ = (3)/(8)\int\limits^7_5 (49y^2-14y^3+y^4)dy \\ \\ = (3)/(8) \left[ (49)/(3) y^3- (14)/(4) y^4+ (1)/(5) y^5\right]^7_5 \\ \\ = (3)/(8) [(5,602.33-8,403.5+3,361.4)-(2,041.67-2,187.5+625)] \\ \\ = (3)/(8) (560.23-479.17)= (3)/(8) (81.06)=30.4`


`V(y)=E(y^2)-[E(y)]^2=30.4-(5.5)^2=30.4-30.25=0.15`



Part C:

Let the required interval be (5, b), then


`P(5\leq y\leq b)= (3)/(4) \\ \\ \Rightarrow \int\limits^b_5 {\left((3)/(8)\right)(7-y)^2} \, dy = (3)/(4) \\ \\ \Rightarrow\int\limits^b_5 {(49-14y+y^2)} \, dy=2 \\ \\ \Rightarrow \left[49y-7y^2+ (1)/(3)y^3\right]^b_5=2 \\ \\ \Rightarrow (49b-7b^2+(1)/(3) b^3)-(245-175+41.67)=2 \\ \\ \Rightarrow (1)/(3) b^3-7b^2+49b-113.67=0 \\ \\ \Rightarrow b=5.74`

Therefore, an interval shorter than (5, 7) in which at least three-fourths of the ph measurements must lie is (5, 5.74).



Part D:


`P(5\ \textless \ Y\ \textless \ 5.5)=\int\limits^(5.5)_5 {\left((3)/(8)\right)(7-y)^2} \, dy \\ \\ = (3)/(8)\int\limits^(5.5)_5 {(49-14y+y^2)} \, dy=(3)/(8)\left[49y-7y^2+ (1)/(3)y^3\right]^(5.5)_5\\ \\ (3)/(8)[(269.5-211.75+55.46)-(245-175+41.67)]=(3)/(8)[113.21-111.67] \\ \\ (3)/(8)(1.54)=0.5775`

Since, the probability that a ph measurement is below 5.5 significant (i.e. 57.75%), we would expect to see a ph measurement below 5.5 very often.