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Question 3 of 13 - Quiz 3
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Approximately 
20% of newborns are born more than 1 week before their due date. A random sample of 20 newborns is selected.
The standard deviation of the sampling distribution for the proportion of your sample that is born more than 7 days before their due date is
0.20 .
3.2 .
0.089 .
0.008 .
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Question Source: baldi se - The Practice of Statistics in The Life Sciences
Publisher:
4:25 PM
4/23/2024

Question 33 of 1313 - Quiz 33\newlineResources\newlineCheck Answer\newlineApproximately 20% 20 \% of newborns are born more than 11 week before their due date. A random sample of 2020 newborns is selected.\newlineThe standard deviation of the sampling distribution for the proportion of your sample that is born more than 77 days before their due date is\newline00.2020 .\newline33.22 .\newline00.089089 .\newline00.008008 .\newlineSearch\newlineQuestion Source: baldi se - The Practice of Statistics in The Life Sciences\newlinePublisher:\newline44:2525 PM\newline44/2323/20242024

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Q. Question 33 of 1313 - Quiz 33\newlineResources\newlineCheck Answer\newlineApproximately 20% 20 \% of newborns are born more than 11 week before their due date. A random sample of 2020 newborns is selected.\newlineThe standard deviation of the sampling distribution for the proportion of your sample that is born more than 77 days before their due date is\newline00.2020 .\newline33.22 .\newline00.089089 .\newline00.008008 .\newlineSearch\newlineQuestion Source: baldi se - The Practice of Statistics in The Life Sciences\newlinePublisher:\newline44:2525 PM\newline44/2323/20242024
  1. Identify Proportion: Identify the proportion of newborns born more than 11 week before their due date.\newlineProportion pp = 2020\% or 0.200.20.
  2. Calculate Standard Deviation: Calculate the standard deviation of the sampling distribution for the proportion using the formula for the standard deviation of a sample proportion: p(1p)/n\sqrt{p(1-p)/n}, where pp is the proportion and nn is the sample size.n=20n = 20 (sample size).
  3. Plug Values and Calculate: Plug the values into the formula and calculate the standard deviation.\newlineStandard deviation = 0.20(10.20)/20=0.20×0.80/20=0.16/20=0.008=0.089\sqrt{0.20(1-0.20)/20} = \sqrt{0.20\times 0.80/20} = \sqrt{0.16/20} = \sqrt{0.008} = 0.089.

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