Question

the length of human pregnancies is approximately normal with mean \\( \mu = 266 \\) days and standard deviation \\( \sigma = 16 \\) days. complete parts (a) through (f). (dont need to use the standard normal distribution table (table z). click here to use the standard normal distribution (page 1).) interpret the probability (select the correct choice below and fill in the answer box within your choice. (round to four decimal places as needed.) \\( \boldsymbol{a} \\). if 100 independent random samples of size \\( n = 7 \\) pregnancies were obtained from this population, we would expect \\( \boldsymbol{\square} \\) sample(s) to have a sample mean of 260 days or less. \\( \boldsymbol{b} \\). if 100 independent random samples of size \\( n = 7 \\) pregnancies were obtained from this population, we would expect \\( \boldsymbol{\square} \\) sample(s) to have a sample mean of exactly 260 days. \\( \boldsymbol{c} \\). if 100 independent random samples of size \\( n = 7 \\) pregnancies were obtained from this population, we would expect \\( \boldsymbol{\square} \\) sample(s) to have a sample mean of 260 days or more. \\( \boldsymbol{10} \\) what is the probability that a random sample of 49 pregnancies has a mean gestation period of 260 days or less? the probability that the mean of a random sample of all pregnancies is less than 260 days is \\( \boldsymbol{\square} \\) (round to four decimal places as needed.)

Explanation:

Step 1: Identify the distribution

The problem states that the length of human pregnancies is approximately normal with mean \(\mu = 266\) days and standard deviation \(\sigma = 16\) days. We are dealing with a normal distribution, and for sample means, we can use the Central Limit Theorem (CLT) which states that the sampling distribution of the sample mean \(\bar{X}\) is normal with mean \(\mu_{\bar{X}}=\mu\) and standard deviation \(\sigma_{\bar{X}}=\frac{\sigma}{\sqrt{n}}\), where \(n\) is the sample size.

Step 2: Calculate the z-score

We want to find the probability that a sample mean of \(n = 4\) pregnancies is less than 250 days. First, we calculate the z-score for the sample mean. The formula for the z-score of a sample mean is:

$$ z=\frac{\bar{x}-\mu_{\bar{X}}}{\sigma_{\bar{X}}}=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}} $$

Substituting the given values: \(\bar{x} = 250\), \(\mu = 266\), \(\sigma = 16\), and \(n = 4\):

$$ z=\frac{250 - 266}{\frac{16}{\sqrt{4}}}=\frac{-16}{\frac{16}{2}}=\frac{-16}{8}=- 2 $$

Step 3: Find the probability from the z-table

We need to find \(P(\bar{X}<250)\), which is equivalent to \(P(Z < - 2)\) where \(Z\) is the standard normal variable. Looking up the z-score of \(-2\) in the standard normal distribution table (or using a calculator with normal distribution functions), we know that the cumulative probability for \(z=-2\) is approximately \(0.0228\).

Answer:

The probability that the mean of a random sample of 4 pregnancies is less than 250 days is \(\boxed{0.0228}\) (or 2.28% when expressed as a percentage).