Calculate the interval [mu_x plusminus 2 sigma_x]. Use the probabilities of part

Question

Calculate the interval [mu_x plusminus 2 sigma_x]. Use the probabilities of part b to find the probability that x will be in this interval. (Round your answers to 4 decimal places. A negative sign should be used instead of parentheses.) Suppose that x is a binomial random variable with n = 5, p = 0.91, and q = .09. For each value of x, calculate p(x). (Round final answers to 4 decimal places.) Find P(x = 3). (Round final answer to 4 decimal places.) Find P(x lessthanorequalto 3). (Do not round intermediate calculations. Round final answer to 4 decimal places.) Find P{x 2). (Do not round intermediate calculations. Round final answer to 4 decimal places.) Use the probabilities you computed in part b to calculate the mean mu_x, the variance, sigma_x^2, and the standard deviation, sigma_x, of this binomial distribution. Show that the formulas for mu_x, sigma_x^2, and sigma_x given in this section give the same results. (Do not round intermediate calculations. Round final answers to mu_x in to 2 decimal places, sigma^2_x and sigma_x in to 4 decimal places.) Calculate the interval [mu_x plusminus 2 sigma_x]. Use the probabilities of part b to find the probability that x will be in this interval.

Explanation / Answer

n=5, p=0.91 and q=0.09

p(0) = 5C0*p^0*q^5 = 1 * 1 * 0.09^5 = 0.00001

p(1) = 5C1*p^1*q^4 = 5 * 0.91 * 0.09^4 = 0.00029

p(2) = 5C2*p^2*q^3 = 10 * 0.91^2 * 0.09^3 = 0.0060

p(3) = 5C3*p^3*q^2 = 10 * 0.91^3 * 0.09^2 = 0.0610

p(4) = 5C4*p^4*q^1 = 5 * 0.91^4 * 0.09^1 = 0.3086

p(5)  = 5C5*p^5*q^0 = 1 * 0.91^5 * 0.09^0 = 0.624

c)

P(x=3) = 0.0610

d)

P(x<=3) = 0.00001+0.00029+0.0060+0.0610 = 0.0674

e)

P(x<3) =P(x<=2) = 0.00001+0.00029+0.0060 = 0.0063

f)

P(x>=4) = 0.3086 + 0.624 = 0.9326

g)

P(x>2) = 0.0610 + 0.3086 + 0.624 = 0.9937

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.