Question

the length of human pregnancies (gestation period) is approximately normally distributed with mean μ = 267 days and standard deviation σ = 17 days.
a) if one pregnancy is randomly selected, what is the probability it lasts less than 262 days?
b) if 20 pregnancies are randomly selected, what is the probability that the sample mean gestation period is 262 days or less?

Explanation:

Part A

Step 1: Identify the distribution and parameters

The length of human pregnancies is normally distributed with mean $\mu = 267$ days and standard deviation $\sigma = 17$ days. We want to find $P(X < 262)$ where $X$ is the length of a single pregnancy.

Step 2: Calculate the z - score

The formula for the z - score is $z=\frac{x-\mu}{\sigma}$. Substituting $x = 262$, $\mu=267$ and $\sigma = 17$ into the formula, we get:
$z=\frac{262 - 267}{17}=\frac{- 5}{17}\approx - 0.2941$

Step 3: Find the probability from the z - table

We want to find $P(Z < - 0.2941)$ where $Z$ is a standard normal variable. Looking up the value of $z=-0.29$ in the standard normal table, the area to the left of $z = - 0.29$ is 0.3859. (For more precise calculation, using a calculator or software, the value is approximately 0.3859)

Step 1: Identify the sampling distribution

The sampling distribution of the sample mean $\bar{X}$ for a sample of size $n = 20$ from a normal population with mean $\mu$ and standard deviation $\sigma$ has mean $\mu_{\bar{X}}=\mu$ and standard deviation $\sigma_{\bar{X}}=\frac{\sigma}{\sqrt{n}}$. Here, $\mu = 267$, $\sigma=17$ and $n = 20$. So $\mu_{\bar{X}}=267$ and $\sigma_{\bar{X}}=\frac{17}{\sqrt{20}}\approx\frac{17}{4.4721}\approx3.799$

Step 2: Calculate the z - score for the sample mean

We want to find $P(\bar{X}<262)$. The z - score for the sample mean is $z=\frac{\bar{x}-\mu_{\bar{X}}}{\sigma_{\bar{X}}}$. Substituting $\bar{x} = 262$, $\mu_{\bar{X}}=267$ and $\sigma_{\bar{X}}\approx3.799$ into the formula, we get:
$z=\frac{262 - 267}{3.799}=\frac{-5}{3.799}\approx - 1.316$

Step 3: Find the probability from the z - table

We want to find $P(Z < - 1.316)$ where $Z$ is a standard normal variable. Looking up the value of $z=-1.32$ (since $-1.316\approx - 1.32$) in the standard normal table, the area to the left of $z=-1.32$ is 0.0934. (Using a calculator or software for more precision, the value is approximately 0.0934, but the given incorrect answer was 0.047, let's recalculate with more precision. $\sigma_{\bar{X}}=\frac{17}{\sqrt{20}}\approx3.7988$. $z=\frac{262 - 267}{3.7988}=\frac{-5}{3.7988}\approx - 1.316$. Looking up $z = - 1.32$ in the standard normal table, the cumulative probability is 0.0934. If we use a calculator, $P(Z < - 1.316)\approx0.0934$)

But if we follow the steps with the given incorrect answer's approach (maybe a miscalculation in the problem's initial wrong answer), let's re - check:

Wait, maybe there was a mistake in the sample size or the standard deviation. Wait, the problem says "20 pregnancies", so $n = 20$. $\sigma_{\bar{X}}=\frac{17}{\sqrt{20}}\approx3.799$. $z=\frac{262 - 267}{3.799}\approx - 1.316$. The correct probability is approximately 0.0934. But if we assume that the user made a mistake in the initial wrong answer (0.047) and we calculate correctly:

The correct z - score calculation:

$\sigma_{\bar{X}}=\frac{17}{\sqrt{20}}\approx3.7988$

$z=\frac{262 - 267}{3.7988}\approx - 1.316$

Using a standard normal calculator, $P(Z < - 1.316)= \Phi(-1.316)$ where $\Phi$ is the cumulative distribution function of the standard normal. $\Phi(-1.316)=1-\Phi(1.316)$. $\Phi(1.316)\approx0.9055$, so $1 - 0.9055 = 0.0945\approx0.094$

But if we consider the initial wrong answer was 0.047, maybe there was a mistake in the problem statement or in the calculation. However, following the correct steps:

Answer:

0.3859

Part B