Question

the lengths of pregnancies in a small rural village are normally distributed with a mean of 260 days and a standard deviation of 17 days. in what range would we expect to find the middle 95% of most lengths of pregnancies? round the answer to one decimal place. between and days > next question

Explanation:

Step1: Recall the property of normal distribution

For a normal - distribution, the middle 95% of the data lies within $z$-scores of $z=- 1.96$ and $z = 1.96$ (approximate values). The $z$-score formula is $z=\frac{x-\mu}{\sigma}$, where $x$ is the value from the data set, $\mu$ is the mean, and $\sigma$ is the standard deviation. We want to solve for $x$. Rearranging the formula gives $x=\mu + z\sigma$.

Step2: Calculate the lower bound

We know that $\mu = 260$, $\sigma=17$, and $z=-1.96$. Substitute these values into the formula $x=\mu + z\sigma$. So, $x_1=260+( - 1.96)\times17=260 - 33.32=226.7$.

Step3: Calculate the upper bound

We use $z = 1.96$, $\mu = 260$, and $\sigma = 17$. Substitute into the formula $x=\mu+z\sigma$. So, $x_2=260 + 1.96\times17=260+33.32 = 293.3$.

Answer:

226.7, 293.3