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The lengths of pregnancies in a small rural village are normally distributed with a mean of 261 days and a standard deviation of 17 days.



In what range would you expect to find the middle 98% of most pregnancies?


Between ___ and ___ .



If you were to draw samples of size 47 from this population, in what range would you expect to find the middle 98% of most averages for the lengths of pregnancies in the sample?


Between ___ and ____ .

Respuesta :

Answer:

(221.39, 300.61) and (255.2223, 266.7777)

Step-by-step explanation:

Given that X, the  lengths of pregnancies in a small rural village are normally distributed with a mean of 261 days and a standard deviation of 17 days

Middle 98% would lie on either side of the mean with probability ±2.33 in the std normal distribution on either side of 0

Corresponding we have x scores as

Between [tex]261-2.33*17[/tex] and [tex]261+2.33*17[/tex]

i.e. in the interval = (221.39, 300.61)

If sample size = 47, then std error of sample would be

[tex]\frac{17}{\sqrt{47} }[/tex]

So 98% of pregnancies would lie between

[tex]261-2.33*\frac{17}{\sqrt{47} }[/tex] and [tex]262+2.33*\frac{17}{\sqrt{47} }[/tex]

= (255.2223, 266.7777)

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