Respuesta :
Answer:
Step-by-step explanation:
Given that the lengths of pregnancies in a small rural village are normally distributed with a mean of 267 days and a standard deviation of 17 days.
X is N(267,17) where x = length of pregnancies in days
For middle 50% we must have on either side 25% area
Hence z= ±0.675
Corresponding x score would be
[tex]267-0.675(17), 267+0.675(15)\\=(255.525,278.475)[/tex]
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Sample size = n =36
Std error of sample = [tex]\frac{17}{\sqrt{36} } =2.83[/tex]
50% would be within
[tex]267-0.675(2.83), 267+0.675(2.83)\\=(265.09,268.91)[/tex]
=between 265.1 and 268.9
Using the normal distribution and the central limit theorem, we have that:
For the population, the middle 50% would be expected to be in the range of 256 to 278 days.
For the samples of 36, the middle 50% would be expected to be in the range of 265 to 269 days.
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Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula, which for a measure X, in a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- The z-score measures how many standard deviations the measure is from the mean.
- Each z-score has an associated p-value, which is the percentile of X.
Population:
- Mean of 267, thus [tex]\mu = 267[/tex].
- Standard deviation of 17, thus [tex]\sigma = 17[/tex].
- The normal distribution is symmetric, which means that the middle 50% is between the 25th percentile and the 75th percentile.
- The 25th percentile is X when Z has a p-value of 0.25, so X when Z = -0.675.
- The 75th percentile is X when Z has a p-value of 0.75, so X when Z = 0.675.
Lower bound:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-0.675 = \frac{X - 267}{17}[/tex]
[tex]X - 267 = -0.675(17)[/tex]
[tex]X = 256[/tex]
Upper bound:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]0.675 = \frac{X - 267}{17}[/tex]
[tex]X - 267 = 0.675(17)[/tex]
[tex]X = 278[/tex]
The middle 50% of pregnancies would be in the range of 256 to 278 days.
Sample of 36:
- By the Central Limit Theorem, the standard deviation of the sampling distribution of sample means of size n is given by [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
- Sample of 36, thus [tex]n = 36, s = \frac{17}{\sqrt{36}} = \frac{17}{6} = 2.8333[/tex]
Lower bound:
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]-0.675 = \frac{X - 267}{2.8333}[/tex]
[tex]X - 267 = -0.675(2.8333)[/tex]
[tex]X = 265[/tex]
Upper bound:
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]-0.675 = \frac{X - 267}{2.8333}[/tex]
[tex]X - 267 = -0.675(2.8333)[/tex]
[tex]X = 269[/tex]
The middle 50% of pregnancies would be in the range of 265 to 269 days.
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