Suppose that the weight of seedless watermelons is normally distributed with mean 6.1 kg. and standard deviation 1.9 kg. Let X be the weight of a randomly selected seedless watermelon. Round all answers to 4 decimal places where possible.a. What is the distribution of X? X ~ N( , )b. What is the median seedless watermelon weight? [] kg.c. What is the Z-score for a seedless watermelon weighing 7.4 kg? []d. What is the probability that a randomly selected watermelon will weigh more than 6.5 kg? []e. What is the probability that a randomly selected seedless watermelon will weigh between 6.4 and 7 kg? []f. The 90th percentile for the weight of seedless watermelons is [] kg.

Answer:a. [tex]X\sim N(\mu = 6.1, \sigma = 1.9)[/tex] b. 6.1 c. 0.6842 d. 0.4166 e. 0.1194 f. 8.5349Step-by-step explanation:a. The distribution of X is normal with mean 6.1 kg. and standard deviation 1.9 kg. this because X is the weight of a randomly selected seedless watermelon and we know that the set of weights of seedless watermelons is normally distributed. b. Because for the normal distribution the mean and the median are the same, we have that the median seedless watermelong weight is 6.1 kg. c. The z-score for a seedless watermelon weighing 7.4 kg is (7.4-6.1)/1.9 = 0.6842 d. The z-score for 6.5 kg is (6.5-6.1)/1.9 = 0.2105, and the probability we are seeking is P(Z > 0.2105) = 0.4166 e. The z-score related to 6.4 kg is [tex]z_{1} = (6.4-6.1)/1.9 = 0.1579[/tex] and the z-score related to 7 kg is [tex]z_{2} = (7-6.1)/1.9 = 0.4737[/tex], we are seeking P(0.1579 < Z < 0.4737) = P(Z < 0.4737) - P(Z < 0.1579) = 0.6821 - 0.5627 = 0.1194 f. The 90th percentile for the standard normal distribution is 1.2815, therefore, the 90th percentile for the given distribution is 6.1 + (1.2815)(1.9) = 8.5349