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If you look at the graph below, the "ceiling" function, is f(x), on that interval of [-4,0], whilst the "floor" function, is g(x)
so, taking that into account, then![\bf \begin{cases} f(x)=x^3+x^2-6x\\\\ g(x)=6x \end{cases}\qquad \displaystyle \int\limits_{-4}^{0}\ \begin{array}{llll} [ceiling]&-&[floor]\\ f(x)&&g(x) \end{array} \\\\\\ thus \\\\\\ \displaystyle \int\limits_{-4}^{0}\ [[x^3+x^2-6x]-[6x]]dx\implies \displaystyle \int\limits_{-4}^{0}\ x^3+x^2-12x \\\\\\ \left. \cfrac{x^4}{4}+\cfrac{x^3}{3}-6x^2 \right]_{0}^{-4}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Bcases%7D%0Af%28x%29%3Dx%5E3%2Bx%5E2-6x%5C%5C%5C%5C%0Ag%28x%29%3D6x%0A%5Cend%7Bcases%7D%5Cqquad%20%5Cdisplaystyle%20%5Cint%5Climits_%7B-4%7D%5E%7B0%7D%5C%20%0A%5Cbegin%7Barray%7D%7Bllll%7D%0A%5Bceiling%5D%26-%26%5Bfloor%5D%5C%5C%0Af%28x%29%26%26g%28x%29%0A%5Cend%7Barray%7D%0A%5C%5C%5C%5C%5C%5C%0Athus%0A%5C%5C%5C%5C%5C%5C%0A%5Cdisplaystyle%20%5Cint%5Climits_%7B-4%7D%5E%7B0%7D%5C%20%5B%5Bx%5E3%2Bx%5E2-6x%5D-%5B6x%5D%5Ddx%5Cimplies%20%0A%5Cdisplaystyle%20%5Cint%5Climits_%7B-4%7D%5E%7B0%7D%5C%20x%5E3%2Bx%5E2-12x%0A%5C%5C%5C%5C%5C%5C%0A%5Cleft.%20%5Ccfrac%7Bx%5E4%7D%7B4%7D%2B%5Ccfrac%7Bx%5E3%7D%7B3%7D-6x%5E2%20%20%5Cright%5D_%7B0%7D%5E%7B-4%7D)
so, taking that into account, then
Let Y denote the maximum age of rabbit.
To calculate probability for Y=a from given table, we need to divide frequency corresponding to y=a with sum of frequencies.
For example P(Y=0) =
Like that P(y=1) = 0.1563, P(Y=2) = 0.2891, P(y=3) = 0.2615, P(Y=4) = 0.1516,
P(Y=5) = 0.0667, P(Y=6) = 0.0391,P(Y=7) = 0.0067, P(Y=8)=0.004 and P(Y=9) =0.
a) P(Y>5) = P(Y=6)+P(Y=7)+P(Y=8)+P(Y=9)
= 0.0391+0.0067+0.004+0 = 0.0498
b) P(2<Y<6) = 0.2615+0.1516+0.0667 = 0.4798
c) P(Y≥3) = 0.2615+0.1516+0.0667+0.0391+0.0067+0.004+0= 0.5296
d)P(Y<6) = 0.02493+0.1563+0.2891+0.2615+0.1516+0.0667 =0.95013
e)If we add (a) and (d), we will get 0.0498+0.95013 = 0.99993≈1
Not surprised,since this is nothing but addition of probabilities for all Y values.
That's we got 1 since numerator and denominator are same.