nment Due Sunday 01.21.18 at 11 45 PM Attempts: 1 Keep the Highest: 1/2 1. An ap

Question

nment Due Sunday 01.21.18 at 11 45 PM Attempts: 1 Keep the Highest: 1/2 1. An application of the sampling distribution of the sample mean People suffering from hypertension, heart disease, or kidney problems may need to limit their intakes of sodium. The public health departmelts in some U.S. states and Canadian provinces require community water systems to notify their customers if the sodium concentration in the drinking water exceeds a designated limit. In Ontario, for example, the notification level is 20 mg/L (milligrams per liter). Suppose that over the course of a particular year the mean concentration of sodium in the drinking water of a water system in Ontario is 18.6 mg/L, and the standard deviation is 6 mg/L Imagine that the water department selects a random sample of 32 water specimens over the course Each specimen is sent to a lab for testing, and at the end of the year the water department computes the mean ration across the 32 specimens. I the mean exceds 20 mg/L, the water department notifies the public and recommends that people who are on sodium-restricted diets inform their physicians of the sodium content in their drinking water. Use the Distributions tool to answer the following questions, adjusting the parameters as necessary Normal Distribution Mean 19.s Standard Deviation 0.85 24 18 20 14 16 10 12 antration of sodium in the drinking water is within the limit, there is a

Explanation / Answer

n=32, =18.6, x = 20 and =6

Thus, Z = (20-18.6)/(6/sqrt(32)) = 1.32

P(sodium over limit) = P(Z >= 1.32) = 0.0934

Thus, z = (x – ) / (/sqrt(n))

Putiing x=20, =18.6, =6,z= 2.32 for 99% right tail

We get n = 100(approx)

Related Questions

Navigate

• Browse (All)
• Browse N
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.