Question 1185732: The lengths of pregnancies in a small rural village are normally distributed with a mean of 270 days and a standard deviation of 17 days.
In what range would you expect to find the middle 95% of most pregnancies?
Between ? and ?
If you were to draw samples of size 30 from this population, in what range would you expect to find the middle 95% 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 by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Here's how to calculate the ranges:
**1. Middle 95% of Individual Pregnancies:**
* For a normal distribution, the middle 95% falls within approximately 1.96 standard deviations of the mean.
* Lower bound: Mean - (1.96 * Standard Deviation) = 270 - (1.96 * 17) = 270 - 33.32 ≈ 236.7
* Upper bound: Mean + (1.96 * Standard Deviation) = 270 + (1.96 * 17) = 270 + 33.32 ≈ 303.3
So, you'd expect to find the middle 95% of pregnancies between approximately **236.7** and **303.3** days.
**2. Middle 95% of Sample Means (n = 30):**
When working with sample means, we use the standard error of the mean (SEM) instead of the population standard deviation. The SEM is calculated as:
SEM = Standard Deviation / √Sample Size = 17 / √30 ≈ 17 / 5.48 ≈ 3.1
* Lower bound: Mean - (1.96 * SEM) = 270 - (1.96 * 3.1) = 270 - 6.076 ≈ 263.9
* Upper bound: Mean + (1.96 * SEM) = 270 + (1.96 * 3.1) = 270 + 6.076 ≈ 276.1
So, you'd expect to find the middle 95% of sample means for pregnancies (with samples of size 30) between approximately **263.9** and **276.1** days.
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