Given that t<span>he 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.](https://tex.z-dn.net/?f=f%28y%29%3D%20%5Cleft%20%5C%7B%20%7B%7B%20%5Cfrac%7B3%7D%7B8%7D%287-y%29%5E2%20%5C%20%5C%20%5C%205%5Cleq%20y%5Cleq7%20%7D%20%5Catop%20%7B0%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20elsewhere%7D%7D%20%5Cright.%20)
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](https://tex.z-dn.net/?f=E%28y%29%3D%20%5Cint%5Climits%5E%5Cinfty_%7B-%5Cinfty%7D%20%7Byf%28y%29%7D%20%5C%2C%20dy%20%5C%5C%20%20%5C%5C%20%3D%20%5Cint%5Climits%5E7_5%20%7By%20%5Cleft%28%5Cfrac%7B3%7D%7B8%7D%5Cright%29%287-y%29%5E2%7D%20%5C%2C%20dy%5C%5C%20%5C%5C%20%3D%20%5Cfrac%7B3%7D%7B8%7D%5Cint%5Climits%5E7_5%20%2849y-14y%5E2%2By%5E3%29dy%20%5C%5C%20%20%5C%5C%20%3D%20%5Cfrac%7B3%7D%7B8%7D%20%5Cleft%5B%20%5Cfrac%7B49%7D%7B2%7D%20y%5E2-%20%5Cfrac%7B14%7D%7B3%7D%20y%5E3%2B%20%5Cfrac%7B1%7D%7B4%7D%20y%5E4%5Cright%5D%5E7_5%20%5C%5C%20%20%5C%5C%20%3D%20%5Cfrac%7B3%7D%7B8%7D%20%5B%281%2C200.5-1%2C600.67%2B600.25%29-%28612.5-583.33%2B156.25%29%5D%20%5C%5C%20%20%5C%5C%20%3D%20%5Cfrac%7B3%7D%7B8%7D%20%28200.08-185.42%29%3D%20%5Cfrac%7B3%7D%7B8%7D%20%2814.66%29%3D5.5)
Part B:
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![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](https://tex.z-dn.net/?f=E%28y%5E2%29%3D%20%5Cint%5Climits%5E%5Cinfty_%7B-%5Cinfty%7D%20%7By%5E2f%28y%29%7D%20%5C%2C%20dy%20%5C%5C%20%0A%5C%5C%20%3D%20%5Cint%5Climits%5E7_5%20%7By%5E2%20%5Cleft%28%5Cfrac%7B3%7D%7B8%7D%5Cright%29%287-y%29%5E2%7D%20%5C%2C%20dy%5C%5C%20%5C%5C%20%3D%20%0A%5Cfrac%7B3%7D%7B8%7D%5Cint%5Climits%5E7_5%20%2849y%5E2-14y%5E3%2By%5E4%29dy%20%5C%5C%20%5C%5C%20%3D%20%5Cfrac%7B3%7D%7B8%7D%20%0A%5Cleft%5B%20%5Cfrac%7B49%7D%7B3%7D%20y%5E3-%20%5Cfrac%7B14%7D%7B4%7D%20y%5E4%2B%20%5Cfrac%7B1%7D%7B5%7D%20y%5E5%5Cright%5D%5E7_5%20%5C%5C%0A%20%5C%5C%20%3D%20%5Cfrac%7B3%7D%7B8%7D%20%5B%285%2C602.33-8%2C403.5%2B3%2C361.4%29-%282%2C041.67-2%2C187.5%2B625%29%5D%20%5C%5C%0A%20%5C%5C%20%3D%20%5Cfrac%7B3%7D%7B8%7D%20%28560.23-479.17%29%3D%20%5Cfrac%7B3%7D%7B8%7D%20%2881.06%29%3D30.4)
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</span><span><span>
![V(y)=E(y^2)-[E(y)]^2=30.4-(5.5)^2=30.4-30.25=0.15](https://tex.z-dn.net/?f=V%28y%29%3DE%28y%5E2%29-%5BE%28y%29%5D%5E2%3D30.4-%285.5%29%5E2%3D30.4-30.25%3D0.15)
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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](https://tex.z-dn.net/?f=P%285%5Cleq%20y%5Cleq%20b%29%3D%20%5Cfrac%7B3%7D%7B4%7D%20%20%5C%5C%20%20%5C%5C%20%5CRightarrow%20%20%5Cint%5Climits%5Eb_5%20%7B%5Cleft%28%5Cfrac%7B3%7D%7B8%7D%5Cright%29%287-y%29%5E2%7D%20%5C%2C%20dy%20%3D%20%5Cfrac%7B3%7D%7B4%7D%20%20%5C%5C%20%20%5C%5C%20%5CRightarrow%5Cint%5Climits%5Eb_5%20%7B%2849-14y%2By%5E2%29%7D%20%5C%2C%20dy%3D2%20%5C%5C%20%20%5C%5C%20%5CRightarrow%20%5Cleft%5B49y-7y%5E2%2B%20%5Cfrac%7B1%7D%7B3%7Dy%5E3%5Cright%5D%5Eb_5%3D2%20%5C%5C%20%20%5C%5C%20%5CRightarrow%20%2849b-7b%5E2%2B%5Cfrac%7B1%7D%7B3%7D%20b%5E3%29-%28245-175%2B41.67%29%3D2%20%5C%5C%20%20%5C%5C%20%5CRightarrow%20%20%5Cfrac%7B1%7D%7B3%7D%20b%5E3-7b%5E2%2B49b-113.67%3D0%20%5C%5C%20%20%5C%5C%20%5CRightarrow%20b%3D5.74%20)
Therefore, </span><span>an interval shorter than (5, 7) in which at least three-fourths of the ph measurements must lie is (5, 5.74).
Part D:
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![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](https://tex.z-dn.net/?f=P%285%5C%20%5Ctextless%20%5C%20Y%5C%20%5Ctextless%20%5C%205.5%29%3D%5Cint%5Climits%5E%7B5.5%7D_5%20%7B%5Cleft%28%5Cfrac%7B3%7D%7B8%7D%5Cright%29%287-y%29%5E2%7D%20%5C%2C%20dy%20%5C%5C%20%20%5C%5C%20%20%3D%20%5Cfrac%7B3%7D%7B8%7D%5Cint%5Climits%5E%7B5.5%7D_5%20%7B%2849-14y%2By%5E2%29%7D%20%5C%2C%20dy%3D%5Cfrac%7B3%7D%7B8%7D%5Cleft%5B49y-7y%5E2%2B%20%5Cfrac%7B1%7D%7B3%7Dy%5E3%5Cright%5D%5E%7B5.5%7D_5%5C%5C%20%5C%5C%20%20%5Cfrac%7B3%7D%7B8%7D%5B%28269.5-211.75%2B55.46%29-%28245-175%2B41.67%29%5D%3D%5Cfrac%7B3%7D%7B8%7D%5B113.21-111.67%5D%20%5C%5C%20%5C%5C%20%5Cfrac%7B3%7D%7B8%7D%281.54%29%3D0.5775)
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.