Mathematics, 25.03.2020 05:32 nyny2129

Exercise 4.50. Suppose an alarm clock has been set to ring after T hours, where T Exp(l/3). Suppose further that your friend has been staring at the clock for exactly 7 hours and can confirm that it has not yet rung. At this point, your friend wants to know when the clock will finally ring. Calculate her conditional probability that she needs to wait at least 3 more hours, given that she has already waited 7 hours. More generally, calculate the conditional probability that she needs to wait at least x more hours, given that she has already waited 7 hours.

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Answer from: cameronchester

a) P(T>10 | T>7) = 0.3678

b) P(T>7+x | T>7) = $e^{\frac{-x}{3} }$

Step-by-step explanation:

a)Conditional Probability that she needs to wait at least 3 more hours, given that she has already waited 7 hours

$T = e^{1/3}$

Three more hours means x = 7+3 = 10

P(T>10 | T>7) = P ( (T>10) ∩ (T>7)) / P (T>7)

P ( (T>10) ∩ (T>7)) = P (T>10)

P (T>10) = 1 - P (T≤10)

P (T≤10) = $1 - e^{\frac{-1}{3} *10}$

P (T≤10) = 0.9643

P (T>10) = 1 - P (T≤10) = 1 - 0.9643

P (T>10) = 0.03567

P (T>7) = 1 - P (T≤7)

P (T≤7) = $1 - e^{\frac{-1}{3} *7}$

P (T≤7) = 0.903

P (T>7) = 1 - P (T≤7) = 1 - 0.903

P (T>7) = 0.097

P(T>10 | T>7) = 0.03567/0.097

P(T>10 | T>7) = 0.3678

b) The conditional probability that she needs to wait at least x more hours, given that she has already waited 7 hours

P(T>7+x | T>7) = P ( (T>7+x) ∩ (T>7)) / P (T>7)

P ( (T>7+x) ∩ (T>7)) = P (T>7+x)

P (T>7+x) = 1 - P (T≤7+x)

P (T≤7+x) = $1 - e^{\frac{-1}{3} *(7+x)}$

P (T>7+x) = 1 - P (T≤(7+x)) = 1 - $(1 - e^{\frac{-1}{3} *(7+x)})$

P (T>7+x) = $e^{\frac{-1}{3} *(7+x)}$

P (T>7+x) = $e^{\frac{-7}{3} } * e^{\frac{-x}{3} }$

P (T>7+x) = $0.097e^{\frac{-x}{3} }$

P (T>7) = 0.097

P(T>7+x | T>7) = $\frac{0.097e^{\frac{-x}{3} }}{0.097}$

P(T>7+x | T>7) = $e^{\frac{-x}{3} }$

Answer from: Quest

you would need to know all the numbers that are on there

step-by-step explanation:

Answer from: Quest

step-by-step explanation:

i found the life tables for the year 2012.

life table of males at age 30. probability of dying between ages x and x+1

total population:

29–30 . . . . . . . . . . . . 0.001030

30–31 . . . . . . . . . . . . 0.001061

males:

29–30 . . . . . . . . . . . . 0.001420

30–31 . . . . . . . . . . . . 0.001448

white population

29–30 . . . . . . . . . . . . 0.001005

30–31 . . . . . . . . . . . . 0.001037

white males

29–30 . . . . . . . . . . . . 0.001372

30–31 . . . . . . . . . . . . 0.001399

non hispanic white population

29–30 . . . . . . . . . . . . 0.001083

30–31 . . . . . . . . . . . . 0.001116

non hispanic white males

29–30 . . . . . . . . . . . . 0.001477

30–31 . . . . . . . . . . . . 0.001503

jacob, non hispanic white male. 30 years old.

probability that jacob lives to be 30 years old.

since the given probabilities are probabilities of dying, i think we simply subtract it from 1 to get the probability of living. we will use the probability of non hispanic white male.

probability of living to be 30 y.o.

29 - 30 : 1 - 0.001477 = 0.998523

30 - 31 : 1 - 0.001503 = 0.998497

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