Question

The average length of a maternity stay in a U.S. hospital is said to be \(2.4\) days with a standard deviation of \(0.9\) days. We randomly survey \(80\) women who recently bore children in a U.S. hospital.

a. In words, \(X= \)

b. In words, \(\bar{X}= \)

c. \(\bar{X} ( , )\)

Short answer

Part a. \(X=\) number of days of maternity stay in U.S. hospital

Part b. \(\bar{X}=\) average length of a maternity stay in a U.S. hospital of \(80\) randomly chosen women who recently bore children.

Part c. \(\bar{X}=N(2.4,0.1)\)

Step-by-step solution

Part a. Step 1. Given information

The average length of a maternity stay in a U.S. hospital is said to be \(2.4\) days with a standard deviation of \(0.9\) days. Randomly survey \(80\) women who recently bore children in a U.S. hospital.

Part a. Step 2. Calculation

The random variable \(X\) is defined by

\(X=\) number of days of maternity stay in U.S. hospital

Hence, the random variable \(X\) is number of days of maternity stay in U.S. hospital.

Part b. Step 1. Calculation

The random variable \(\bar{X}\) is defined by

\(\bar{X}=\) average length of a maternity stay in a U.S. hospital of \(80\) randomly chosen women who recently bore children.

Hence, the random variable \(\bar{X}\) is the average length of a maternity stay in a U.S. hospital of \(80\) randomly chosen women who recently bore children.

Part c. Step 1. Calculation

The distribution of \(\bar{X}\) is Normal.

\(\bar{X}=N(\mu_{x},\frac{\sigma_{x}}{\sqrt{n}}\)

\(\mu_{x}=2.4\)

\(\sigma_{x}=0.9\)

\(n=80\)

Therefore, \(\bar{X}=N(\mu_{x},\frac{\sigma_{x} {\sqrt{n}}= \bar{X}=N(\2.4,\frac{0.9}{\sqrt{80}}= \bar{X}=N(2.4,0.1)\)

Hence, \(\bar{X}=N(2.4,0.1)\)