Problem 1 is posted bellow to refer to. Thank you please show all work. [10 poin
ID: 3307359 • Letter: P
Question
Problem 1 is posted bellow to refer to. Thank you please show all work. [10 points] Refer to Problem 1. . (a) Suppose that we define an individual who has automobile insurance from the company as a success" if they have no moving violations. Calculate the probability that, in a randonm sample of 10 individuals, exactly 4 will have no moving violations. [3 points) (b) Suppose that we define an individual as a "success" if they have at least two moving violations. Calculate the probability that at most 3 out of 10 pcople will each have at least two moving violations. [3 points] (c) Calculate the (i) mean and the (ii) standard deviation of the number of individuals out of 10 who have exactly one moving violation. [2 points cach]Explanation / Answer
a. There are two outcomes-success (if no moving violations) and failure (if moving violations). There are n=10 independent, random trials, with probability of success, p=0.50 (no moving violations). The probability of success is constant throughout the trials.This accounts for binomial distribution, hence use P(X,r)=nCr(p)^r(1-p)^n-r, where, p denotes probability of success, r denotes specific number of success in n trials.
P(X=4)=10C4(0.5)^4(1-0.5)^6=0.2051
b. The probability of success, p=0.3 (0.15+0.10+0.05) [it is said probability of success refer to probability of atleast 2 moving violations, and atleast 2 means cumulative probability comprising 2, 3, 4 moving violations]
It is required to find P(X<=3). Use, binomial distribution table, to locate n=10, x=3, p=0.3. The require dprobability is 0.650. The table gives cumulative probability P(X<=x). One can manually solve this by applying formula.
P(X<=3)=P(X=0)+P(X=1)+P(X=2)+P(X=3)=0.0282+0.1211+0.2335+0.2668=0.6496~0.650
c. Mean, mu=np=10*0.20=2 [note, p=0.20], standard deviation, sigma=sqrt np(1-p)=sqrt 10*0.20*0.80=1.26
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