Question

Oxygen demand is a term biologists use to describe the oxygen needed by fish and other aquatic organisms for survival. The Environmental Protection Agency conducted a study of a wetland area. In this wetland environment, the mean oxygen demand was μ = 9.8 mg/L with 95% of the data ranging from 6.0 mg/L to 13.6 mg/L. Let x be a random variable that represents oxygen demand in this wetland environment. Assume x has a probability distribution that is approximately normal. (a) Use the 95% data range to estimate the standard deviation for oxygen demand. (b) An oxygen demand below 8 indicates that some organisms in the wetland environment may be dying. What is the probability that the oxygen demand will fall below 8 mg/L? (Round your answer to four decimal places.) (c) A high oxygen demand can also indicate trouble. An oxygen demand above 12 may indicate an overabundance of organisms that endanger some types of plant life. What is the probability that the oxygen demand will exceed 12 mg/L? (Round your answer to four decimal places.)

Answer 1

Answer:

Given that:

The Environmental Protection Agency conducted a study of a wetland area. In this wetland environment, the mean oxygen demand was μ = 9.8 mg/L with 95% of the data ranging from 6.0 mg/L to 13.6 mg/L. Let x be a random variable that represents oxygen demand in this wetland environment.

a) Use the 95% data range to estimate the standard deviation for oxygen demand

According to the 68-95-99 rules, 95% of the data lies within two standard deviations of the mean.
Hence,
- 2*s = 6
9.8 - 2s = 6
3.8 = 2s
s = 1.9 mg/L

b) An oxygen demand below 8 indicates that some organisms in the wetland environment may be dying. What is the probability that the oxygen demand will fall below 8 mg/L?

P(X < 8) = P(z < (8-9.8)/1.9)

= P(z < -0.95)

= 0.1711

c)  A high oxygen demand can also indicate trouble. An oxygen demand above 12 may indicate an overabundance of organisms that endanger some types of plant life. What is the probability that the oxygen demand will exceed 12 mg/L?

P(X > 12) = P(z > (12-9.8)/1.9)

= P(z > 1.16)

= 1 - P(z < 1.16)

= 1 - 0.8770

= 0.1230