Assume that a procedure yields a binomial distribution with a trial repeated n=5n=5 times. Use some form of technology to find the cumulative probability distribution given the probability p=0.227p=0.227 of success on a single trial. Assume that a procedure yields a binomial distribution with a trial repeated n = 5 times. Use some formof technology to find the cumulative probability distribution given the probability p = 0.227 of success ona single trial.(Report answers accurate to 4 decimal places.)P(X < k)k

Name: Assume that a procedure yields a binomial distribution with a trial re
Uploaded: 2024-03-20T07:06:41.000Z
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Description: Assume that a procedure yields a binomial distribution with a trial repeated n=5n=5 times. Use some form of technology to find the cumulative probability distribution given the probability p=0.227p=0.227 of success on a single trial. Assume that a pr

Given information: Trials, n=5 Probability of success on a single trial, p=0.227The probability density function of a random variable X following binomial distribution is: PX=x=Cxn px 1-pn-x Case 1: k=0 PX≤k=PX≤0=PX=0=C05*0.2270*1-0.2275-0=0.2760 Therefore, the required probability is 0.2760. Case 2: k=1 PX≤k=PX≤1=PX=0+PX=1=C05*0.2270*1-0.2275-0+C15*0.2271*1-0.2275-1=0.6812 Therefore, the required probability is 0.6812. Case 3: k=2 PX≤k=PX≤2=PX=0+PX=1+P(X=2)=C05*0.2270*1-0.2275-0+C15*0.2271*1-0.2275-1+C25*0.2272*1-0.2275-2=0.9192 Therefore, the required probability is 0.9192. Case 4: k=3 PX≤k=PX≤3=PX=0+PX=1+P(X=2)+P(X=3)=C05*0.2270*1-0.2275-0+C15*0.2271*1-0.2275-1+C25*0.2272*1-0.2275-2+C35*0.2273*1-0.2275-3=0.9891 Therefore, the required probability is 0.9891. Case 5: k=4 PX≤k=PX≤4=PX=0+PX=1+P(X=2)+P(X=3)+P(X=4)=C05*0.2270*1-0.2275-0+C15*0.2271*1-0.2275-1+C25*0.2272*1-0.2275-2+C35*0.2273*1-0.2275-3+C45*0.2274*1-0.2275-4=0.9994 Therefore, the required probability is 0.9994. Case 6: k=5 PX≤k=PX≤5=PX=0+PX=1+P(X=2)+P(X=3)+P(X=4)=C05*0.2270*1-0.2275-0+C15*0.2271*1-0.2275-1+C25*0.2272*1-0.2275-2+C35*0.2273*1-0.2275-3+C45*0.2274*1-0.2275-4+C55*0.2275*1-0.2275-5=1 Therefore, the required probability is 1.Therefore, the calculated probabilities can be summarized in the following table: k P(X≤k) 0 0.2760 1 0.6812 2 0.9192 3 0.9891 4 0.9994 5 1