The lifespan of females in certain part of the world have been described using a
ID: 3352619 • Letter: T
Question
The lifespan of females in certain part of the world have been described using a continous probability model. The probability that a selected female lives beyond the age x is 3. exp 7688 What is the probability that (a) a randomly selected female does not live past the age of 75 (b) what is the probability that a female lives past 95 years (c) knowing that the people in a certain community live past 80 years, what is the probability that a female in that community will live past 95 years. A student comes to the class late 30% of the time. If the class meets 5 times a week, (a) find the probability that the student is late for at least for 3 classes in a week and (b) the probability that the student is not late at all in any given week 4.Explanation / Answer
3)
P =e^(-x^2/7688)
a) P(x<75) =1-P(x>75)
=1-e^(-75^2/7688)
=1-e^(-0.731659)
=1-0.4811
=0.5189
b) P(x> 95) =e^(-95^2/7688)
= e^(-9025/7688)
=e^(-1.1739)
=0.3092
c) P(x>80) =e^(-80^2/7688)
=e^(-6400/7688)
=e^(-0.83246)
=0.4350
P( female live past 95 given that they live past80) = P(x>95) /P(x>80)
= 0.3092/0.4350
=0.7108
4)
let p = probability that student is late
p=0.3
n=5
q=1-p=1-0.3 =0.7
We need to use binomial distribution formula
P(x=r) =nCr p^r q^(n-r)
a) P(x>= 3) = P(x=3) +P(x=4) +P(x=5)
=5C3 (0.3)^3 (0.7)^2 +5C4 (0.3)^4 (0.7)^1 +5C5 (0.3)^5
=0.1323+0.02835 +0.00243
=0.16308
b) P(x=0) =5C0 (0.3)^0 (0.7)^(5-0)
=1 *1 *0.7^5
=0.7^5 =0.16807
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