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

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

Part A:

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



Part B:

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

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



Part C:

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

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

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:

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

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.