Question

Here are some (simulated) data on the maximum age distribution in rabbits:

age: 0 1 2 3 4 5 6 7 8 9
frequency: 37 232 429 388 225 99 58 10 6 0 Give the following:
a) Pr(Y > 5) b) Pr(2 < Y < 6) c) Pr(Y ≥ 3) d) Pr(Y < 6)
e) Add (a) and (d). Are you surprised? Why or why not?

Answer 1

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) =
`(37)/(37+232+429+388+225+99+58+10+6+0) =(37)/(1484) = 0.02493`

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.