Question

ch 15 an srs of 25 recent birth records at the local hospital was selected. in the sample, the average birth weight was $\bar{x}=119.6$ ounces. suppose the standard deviation is known to be $sigma = 6.5$ ounces. assume that in the population of all babies born in this hospital, the birth weights follow a normal distribution, with mean $mu$. the standard deviation of the sampling distribution of the mean is
6.52 ounces.
1.30 ounces.
0.02 ounce.
0.38 ounce.

question 14
ch 15 suppose youre in a class of 35 students. the instructor takes a simple random sample of seven students and observes their heights. imagine all of the different samples possible. let x denote the tallest height in your sample. the distribution of all values taken by x in all possible samples of seven students selected from the 35 students in your class is
the sampling distribution of x.
the standard deviation of values.
the parameter.
the probability that x is obtained.

Explanation:

Step1: Recall the formula for standard deviation of the sampling - distribution of the mean

The formula for the standard deviation of the sampling - distribution of the mean (also known as the standard error) is $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$, where $\sigma$ is the population standard deviation and $n$ is the sample size.

Step2: Identify the values of $\sigma$ and $n$

We are given that $\sigma = 6.5$ ounces and $n = 25$.

Step3: Calculate the standard deviation of the sampling - distribution of the mean

Substitute the values into the formula: $\sigma_{\bar{x}}=\frac{6.5}{\sqrt{25}}=\frac{6.5}{5}=1.3$ ounces.

For the second question:
The distribution of all values taken by a statistic (in this case, $X$ which is the tallest height in the sample) in all possible samples of a given size from a population is called the sampling distribution of that statistic.

Answer:

Question 13: 1.30 ounces
Question 14: the sampling distribution of X