Question

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. In what range would you expect to find the middle 50% of most pregnancies? Between and . If you were to draw samples of size 36 from this population, in what range would you expect to find the middle 50% of most averages for the lengths of pregnancies in the sample? Between and . Enter your answers as numbers. Your answers should be accurate to 1 decimal places.

Answer 1

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

$267-0.675(17), 267+0.675(15)\\=(255.525,278.475)$

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Sample size = n =36

Std error of sample = $\frac{17}{\sqrt{36} } =2.83$

50% would be within

$267-0.675(2.83), 267+0.675(2.83)\\=(265.09,268.91)$

=between 265.1 and 268.9

Answer 2

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 $\mu$ and standard deviation $\sigma$, is given by:

$Z = \frac{X - \mu}{\sigma}$

• 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 $\mu = 267$.
• Standard deviation of 17, thus $\sigma = 17$.

• 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:

$Z = \frac{X - \mu}{\sigma}$

$-0.675 = \frac{X - 267}{17}$

$X - 267 = -0.675(17)$

$X = 256$

Upper bound:

$Z = \frac{X - \mu}{\sigma}$

$0.675 = \frac{X - 267}{17}$

$X - 267 = 0.675(17)$

$X = 278$

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 $s = \frac{\sigma}{\sqrt{n}}$.
• Sample of 36, thus $n = 36, s = \frac{17}{\sqrt{36}} = \frac{17}{6} = 2.8333$

Lower bound:

$Z = \frac{X - \mu}{s}$

$-0.675 = \frac{X - 267}{2.8333}$

$X - 267 = -0.675(2.8333)$

$X = 265$

Upper bound:

$Z = \frac{X - \mu}{s}$

$-0.675 = \frac{X - 267}{2.8333}$

$X - 267 = -0.675(2.8333)$

$X = 269$

The middle 50% of pregnancies would be in the range of 265 to 269 days.

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