Question

question 7
a study reported that the length of pregnancy from conception to birth is approximately normally distributed with mean \\( \mu=272 \\) days and standard deviation \\( \sigma=9 \\) days. what proportion of pregnancies last longer than 280 days? use normal dist

Explanation:

Step1: Calculate the z - score

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(x = 280\), \(\mu=272\), and \(\sigma = 9\).

$$z=\frac{280 - 272}{9}=\frac{8}{9}\approx0.89$$

Step2: Find the proportion using the standard normal table

We want to find \(P(X>280)\), which is equivalent to \(P(Z > 0.89)\) since \(X\) (pregnancy length) is normally distributed and \(Z=\frac{X-\mu}{\sigma}\).
We know that \(P(Z>z)=1 - P(Z\leq z)\). From the standard - normal table, \(P(Z\leq0.89) = 0.8133\).

$$P(Z>0.89)=1 - 0.8133=0.1867$$

Answer:

The proportion of pregnancies that last longer than 280 days is approximately \(0.1867\)