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Scientists at the Hopkins Memorial Forest in western Massachusetts have been collecting meteorological and environ- mental data in the forest data for more than 100 years. In the past few years, sulfate content in water samples from Birch Brook has averaged 7.48 mg/L with a standard deviation of 1.60 mg/L. (a) What is the standard error of the sulfate in a collection of 10 water samples? (b) If 10 students measure the sulfate in their samples, what is the probability that their average sulfate will be between 6.49 and 8.47 mg/L? (c) What do you need to assume for the probability calculated in (b) to be accurate?

Respuesta :

Answer:

(a) standard error of the sulfate in a collection of 10 water samples = .506 mg/L

( b) the probability that their average sulfate will be between 6.49 and 8.47 mg/L = .950 mg/L

(c) We need to assume sample is random and normal sampling distribution probability calculated in (b) to be accurate.

Step-by-step explanation:

Given -

Average sulfate content [tex](\nu )[/tex] = 7.48 mg/L

Standard deviation [tex](\sigma )[/tex] = 1.60 mg/L

(a) standard error of the sulfate in a collection of 10 water samples =

  standard error = [tex]{\frac{\sigma}{\sqrt{n}}}[/tex]  = [tex]{\frac{1.60}{\sqrt{10}}}[/tex]   = .506 mg/L

(b) If 10 students measure the sulfate in their samples, what is the probability that their average sulfate will be between 6.49 and 8.47 mg/L =

[tex]P(6.49\leq \overline{X}\leq 8.47)[/tex] = [tex]P(\frac{6.49 - 7.48 }{\frac{1.60}{\sqrt{10}}}\leq \frac{\overline{X} - \nu }{\frac{\sigma}{\sqrt{n}}}\leq \frac{8.47 - 7.48 }{\frac{1.60}{\sqrt{10}}})[/tex]

                               = [tex]P(-1.96\leq Z \leq 1.96)[/tex]

                               = [tex]P( Z\leq 1.96) - P( Z\leq -1.96)[/tex]

                               = .975 - .025 = .950 mg/L

(c) We need to assume sample is random and normal sampling distribution probability calculated in (b) to be accurate.

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