Question

The salinity, or salt content, in the ocean is expressed in parts per thousand (ppt). The number varies with depth, rainfall, evaporation, river runoff, and ice formation. The mean salinity of the oceans is 35 ppt. Suppose the distribution of salinity is normal and the standard deviation is 0.53 ppt, and suppose a random sample of ocean water from a region in a specific ocean is obtained. (a) What is the probability that the salinity is more than 36 ppt? (Round your answer to four decimal places) 0.0296 (b) What is the probability that the salinity is less than 33.5 ppt? (Round your answer to four decimal places.) 0.0023 (C) A certain species of fish can only survive if the salinity is between 33 and 35 ppt. What is the probability that this species can survive in a randomly selected area? (Round your answer to four decimal places.) 0 4999 in this interval, (Round your answers to four (a) Find a symmetric interval about the mean salinity such that 50% of all salinity levels decimal places.) (325x35x)

Answer 1

Given that, u = 35,0 =0.53 For 50% around mean should be lies between 25% and 75% values The z values corresponding to 25% and 75% are, Z = +0.6745 Therefore, the symmetrical interval about the mean salinity such that 50% of all salinity levels lie in this interval is, u$:(0) 35+0.6745(0.53) (35–0.3575,35+0.3575) (34.6425,35.3575) Answer: (34.6425, 35.3575)