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 0.51 ppt, and suppose a random sample of ocean water from a region in a specific ocean is obtained.

What is the probability that the salinity is more than 36 ppt? (Round your answer to four decimal places.)

_____________

What is the probability that the salinity is less than 33.5 ppt? (Round your answer to four decimal places.)

____________

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.)

____________

Find a symmetric interval about the mean salinity such that 50% of all salinity levels lie in this interval. (Round your answers to four decimal places.)

_________ , __________

Answer 1

Solution :

Given that ,

mean = = 35

standard deviation = = 0.51

a) P(x > 36) = 1 - p( x< 36)

=1- p P[(x - ) / < (36 - 35) / 0.51 ]

=1- P(z < 1.96)

Using z table,

= 1 - 0.9750

= 0.0250

b) P(x < 33.5) = P[(x - ) / < (33.5 - 35) / 0.51]

= P(z < -2.94)

Using z table,

= 0.0016

c) P(33 < x < 35) = P[(33 - 35)/ 0.51) < (x - ) /  < (35 - 35) / 0.51) ]

= P(-3.92 < z < 0)

= P(z < 0) - P(z < -3.92)

Using z table,

= 0.5000 - 0

= 0.5000

d) Using standard normal table,

P( -z < Z < z) = 50%

= P(Z < z) - P(Z <-z ) = 0.50

= 2P(Z < z) - 1 = 0.50

= 2P(Z < z) = 1 + 0.50

= P(Z < z) = 1.50 / 2

= P(Z < z) = 0.75

= P(Z < 0.67) = 0.75

= z  ± 0.67

Using z-score formula,

x = z * +

x = -0.67 * 0.51 + 35

x = 34.6583

Using z-score formula,

x = z * +

x = 0.67 * 0.51 + 35

x = 35.3417

The middle 50% are from 34.6583 to 35.3417