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elixir [45]
2 years ago
7

A regional transportation authority is interested in estimating the mean number of minutes working adults in the region spend co

mmuting to work on a typical day. A random sample of working adults will be selected from each of three strata: urban, suburban, and rural. Selected individuals will be asked the number of minutes they spend commuting to work on a typical day. Why is stratification used in this situation?
A. To remove bias when estimating the proportion of working adults living in urban, suburban, and rural areas.B. To remove bias when estimating mean commuting time.C. To reduce bias when estimating mean commuting time.D. To decrease the variability in estimates of the proportion of working adults living in urban, suburban, and rural areas.E. To decrease the variability in estimates of the mean commuting time.
Mathematics
1 answer:
garri49 [273]2 years ago
4 0

Answer:

To reduce bias when estimating mean commuting time

Step-by-step explanation:

Given that the selection of adults is based on the three strata found in the region; urban, suburban, and rural areas where the population and proportion of adults commuting to work is progressively decreasing, using a uniform number or sample size will not increase sample error as there are more commuters in the urban areas than in the suburban or rural areas, likewise, there are more commuters in the suburban area than in the rural areas, and as such the proportion of those sampled in the urban  areas should be more than those surveyed in the suburban while those surveyed in the suburban areas should be more than those surveyed in the rural areas to reduce bias when estimating the mean as there will be more general representation.

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Suppose that the weight of navel oranges is normally distributed with a mean µµ = 8 ounces, and a standard deviation σσ = 1.5 ou
monitta

Answer:

Hello some parts of your question is missing below is the missing part

c. If you randomly select a navel orange, what is the probability that it weighs between6.2 and 7 ounces

Answer: A) 0.0099

              B) 0.6796

              C) 0.13956

Step-by-step explanation:

weight of Navel oranges evenly distributed

mean ( u ) = 8 ounces

std ( б )= 1.5

navel oranges = X

A ) percentage of oranges weighing more than 11.5 ounces

P( x > 11.5 ) = P ( \frac{x - u}{ std} > \frac{11.5-8}{1.5} )

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B) percentage of oranges weighing less than 8.7 ounces

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                    = P ( Z < 0.4667 ) = 0.6796

                    = 67.96%

C ) probability of orange selected weighing between 6.2 and 7 ounces?

P ( 6.2 < X < 7 ) = P (\frac{6.2-8}{1.5} <  \frac{x - u}{ std} < \frac{7-8}{1.5} )

                          = P ( -1.2 < Z < -0.66 )

                          = Ф ( -0.66 ) - Ф(-1.2) = 0.13956

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