SOLUTION: a) A random variable x has a binomial distribution with n = 12. Given that p = 0.25, find (i) p (x < 5) (From table III or IV of Appendix A) (ii) p (x ≥ 7) (From table II

Question 1198592: a) A random variable x has a binomial distribution with n = 12.
Given that p = 0.25, find
(i) p (x < 5) (From table III or IV of Appendix A)
(ii) p (x ≥ 7) (From table III or IV of Appendix A)
b)Given that p(x=0) = 0.05, find the value of p to 3 decimal places.
Answer by ElectricPavlov(122) (Show Source): You can put this solution on YOUR website!
**a) Binomial Distribution with n = 12 and p = 0.25**
* **(i) p(x < 5)**
* Using a binomial probability table or calculator (like the one provided in the code response), we find:
* p(x < 5) = 0.8819
* **(ii) p(x ≥ 7)**
* p(x ≥ 7) = 1 - p(x < 7)
* Using a binomial probability table or calculator, we find:
* p(x < 7) = 0.9856
* Therefore, p(x ≥ 7) = 1 - 0.9856 = 0.0144
**b) Finding the value of p given p(x=0) = 0.05**
* We know that for a binomial distribution:
* p(x=0) = (nCx) * p^x * (1-p)^(n-x)
* where nCx is the binomial coefficient (number of combinations of n items taken x at a time)
* In this case, n = 12 and x = 0
* p(x=0) = (12C0) * p^0 * (1-p)^(12-0)
* p(x=0) = 1 * 1 * (1-p)^12
* We are given p(x=0) = 0.05
* Therefore, 0.05 = (1-p)^12
* Solve for p:
* p = 1 - (0.05)^(1/12)
* p ≈ 0.221
**Therefore:**
* **a) (i) p(x < 5) = 0.8819**
* **(ii) p(x ≥ 7) = 0.0144**
* **b) The value of p is approximately 0.221**

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