A randomly selected 90 years old man has a 80% chance survive the year. A random

Question

A randomly selected 90 years old man has a 80% chance survive the year. A random sample of 35 individuals in the 90-years old man population was obtained. Q2. What is the probability that 20 to 30 of them will survive the year? 1) 0.9598 2) 0.8560 3) 0.9638 4) 0.9422 5) 0.9087 Q3. What is the probability that at least 25 of them will survive the year? 5) 0.9253 1) 0.6137 2) 0.8001 3) 0.8718 4) 0.7125 Suppose that the number of traffic accidents near the campus has a Poisson distribution with an average of 1.5 in every 3 hours. In a similar traffic condition, Q4. what is the probability that the number of traffic accidents will exceed 2 per hour? 5) 0.0120 1) 0.0170 2) 0.0098 3) 0.0200 4) 0.0144 Q5. Be within 2 S.D. of the mean value per hour? 3) 0.9098 4) 0.9197 5) 0.8995 1) 0.9293 2) 0.8889 The number of hits to a Web site follows a Poisson process. Hits occur at the rate of 3.8 per minute between 8:00 P.M. and 10:30 P.M. Q6. What is the probability that there will be in between 15 and 25, inclusively, hits between 8:25 P.M. and 8:30 P.M.? 1) 0.6939 2) 0.7661 3) 0.7473 5) 0.6194 Q7. What is the probability that there will be in between 15 and 25, inclusively, hits between 9:25 P.M. and 9:35 P.M.? 1) 0.1225 2) 0.0673 3) 0.0033 4) 0.0345 5) 0.0166 Form D pg. 1

Explanation / Answer

Q2) since this question can be solved by binomial theorm where we need to find the probability of a person living with age>90 years

P(X=r) = nCr p^r q^(n-r) equation of binomial

so adjusting the same for our question we have r from 20 to 30, p=0.8,q=0.2 (i.e. 1-p),n=35

P(X>=20,X<=30)=P(X=20)+P(X=21)+---P(X=30) to solve this we need the help of a software and so I have used r to solve the same.

R code given below:

> sum(dbinom(20:30,35,0.8))
[1] 0.8559835

So the answer is 2)0.8560

Q3) Same way now we need to calculate P(X>=25) for atleast 25 people to survive

modifying the r code accordingly we get

> sum(dbinom(25:35,35,0.8))
[1] 0.925291

ans = 5)0.9253

Q4) now these questions are of poission distribution

The formula for the same is:

P(x; ) = (e-) (x) / x! where x is the actual number of successes that result from the experiment, and e is approximately equal to 2.71828.

so again here also we ned to take the help of r to solve the question

> 1-sum(dpois(0:2,1.5/3)) [removing the probabilities of 0 to 2 accidents from 1 to get the ans]
[1] 0.01438768

so the answer is

4)0.0144

As per the Chegg policy I have explained the first set of questions here. Can't go on for the next que as it will take time and can't be solved in the stipulated time. Hope you understand. If you want the ans for the next also pls repost part 2.Hope you understand.

Pls upvote the ans if it has really helped you. Good Luck!!

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