x You received no credit for this question in the previous attempt. View previous attempt 19 If p= 1 and n = 5, then the corresponding binomial distribution is Multiple Choice eBook Print References O bimodal O left skewed symmetric O right skewedSuppose that x is a binomial random variable with n = 5, p = .70, and q= .30. (b) For each value of x, calculate p(x). (Round final answers to 4 decimal places.) p(0) = 0.0024 , P(1) = 0.0284 , P(2) = 0. 1323 , P(3) = 0.3087 p(4) = 0.3602 , p(5) = 0. 1681 (c) Find P(x = 3). (Round final answer to 4 decimal places.) P(x = 3) 0.3087 You received partial credit in the previous attempt. Suppose that xis a binomial random variable with n = 5, p= .70, and q= .30. (b) For each value of x, calculate p(x). (Round nal answers to 4 decimal places.) _-_-- (c) Find P(x= 3). (Round final answer to 4 decimal places.) P(x = 3) 0.3087 (d) Find P(x 4). (Do not round intermediate calculations. Round final answer to 4 decimal places.) P(x 2 4) 0.5282(g) Find P(x > 2). (Do not round intermediate calculations. Round final answer to 4 decimal places.) P(X > 2) 0.9692 (h) Use the probabilities you computed in part b to calculate the mean , the variance, or , and the standard deviation, o,, of this binomial distribution. Show that the formulas for My , ox , and ox given in this section give the same results. (Do not round intermediate calculations. Round final answers to My in to 2 decimal places, ox and Ox in to 4 decimal places.) ux 3.50 ox^2 1.05 OX 1.0247(i) Calculate the interval [My + 20 ]. 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.) The interval is [ 1.4506 5.5494 ] P( 1.4506 SXS 5.5494 ) = 0.8012e You received partial credit in the previous attempt. The United States Golf Association requires that the weight ofa golf ball must not exceed 1.62 oz. The association periodically checks golf balls sold in the United States by sampling specic brands stocked by pro shops. Suppose that a manufacturer claims that no more than 18 percent of its brand of golf balls exceed 1.62 oz. in weight. Suppose that 24 of this manufacturer's golf balls are randomly selected, and let x denote the number of the 24 randomly selected golf balls that exceed 1.62 oz. Refer to the Binomial table given below. Excel Output of the Binomial Distribution with n = 24, p= .18, and q = .82 Binomial distribution with n= 24 and p = .18 P(X = x) 0.0085 0.0450 0.1136 0.1829 0.2107 0.1850 [ WIPDJNHON (a) Find P(x= 0), that is, nd the probability that none of the randomly selected golf balls exceeds 1.62 oz. in weight. (Use table values rounded to 4 decimal places for calculations. Round your answer to 4 decimal places.) m (b) Find the probability that at least one of the randomly selected golf balls exceeds 1.62 oz. in weight. (Use table values rounded to 4 decimal places for calculations. Round your answer to 4 decimal places.) P(x 2 1) 0.7735 (c) Find P(X $ 3). (Use table values rounded to 4 decimal places for calculations. Round your answer to 4 decimal places.) P(x 5 3) 0.9474 (d) Find P(x 2 2). (Use table values rounded to 4 decimal places for calculations. Round your answer to 4 decimal places.) l P(x 2 2) (e) Suppose that 2 of the 24 randomly selected golf balls are found to exceed 1.62 02. Using your result from part d, do you believe the claim that no more than 18 percent of this brand of golf balls exceed 1.62 oz. in weight? _ . the probability of this result is m if the claim is true