reasoning the table shows the daily changes in the barometric pressure (in inches of mercury) for four…

reasoning the table shows the daily changes in the barometric pressure (in inches of mercury) for four days.\n| day | change in pressure |\n| ---- | ---- |\n| monday | -0.05 |\n| tuesday | 0.09 |\n| wednesday | -0.04 |\n| thursday | -0.08 |\n\na. what is the mean change?\n\nb. the mean change for monday through friday is -0.01 inch. what is the change in the barometric pressure on friday?

reasoning the table shows the daily changes in the barometric pressure (in inches of mercury) for four days.\n| day | change in pressure |\n| ---- | ---- |\n| monday | -0.05 |\n| tuesday | 0.09 |\n| wednesday | -0.04 |\n| thursday | -0.08 |\n\na. what is the mean change?\n\nb. the mean change for monday through friday is -0.01 inch. what is the change in the barometric pressure on friday?

Answer

Explanation:

Step1: Recall the mean formula

The mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$, where $x_{i}$ are the data - points and $n$ is the number of data - points.

Step2: Identify the data for part a

We have data for 4 days: $x_1=-0.05$, $x_2 = 0.09$, $x_3=-0.04$, $x_4=-0.08$, and $n = 4$.

Step3: Calculate the sum for part a

$\sum_{i = 1}^{4}x_{i}=-0.05 + 0.09-0.04-0.08=-0.08$.

Step4: Calculate the mean for part a

$\bar{x}=\frac{-0.08}{4}=-0.02$.

Step5: Set up an equation for part b

Let the change on Friday be $x$. We know that $n = 5$ (Monday - Friday) and the mean $\bar{x}=-0.01$. The sum of the first 4 values is $-0.08$ (from Step 3). Using the mean formula $\bar{x}=\frac{-0.08 + x}{5}$.

Step6: Solve the equation for part b

Since $\bar{x}=-0.01$, we have $-0.01=\frac{-0.08 + x}{5}$. Cross - multiply: $-0.01\times5=-0.08 + x$. So, $-0.05=-0.08 + x$. Then $x=-0.05 + 0.08=0.03$.

Answer:

a. $-0.02$ b. $0.03$