Question

The daily high temperatures, in degrees Celsius, for two weeks were recorded. The first week had a mean high temperature of 8 degrees Celsius and a mean absolute deviation (MAD) of 1 degree Celsius. Here is a list of the high temperatures, in degrees Celsius, for the second week. A. What is the mean temperature for the second week? B. What is the mean absolute deviation for the second week? C. Which week had greater variability in high temperature? Explain your reasoning. (When necessary, round all answers to the nearest hundredths place. Type all answers below.

Answer 1

(A) The mean temperature for the second week is 4.86 degrees Celsius.

(B) The mean absolute deviation for the second week is 4.86 degrees Celsius.

(C) The second week had greater variability in high temperature.

Explanation:

We are given that the daily high temperatures, in degrees Celsius, for two weeks were recorded. The first week had a mean high temperature of 8 degrees Celsius and a mean absolute deviation (MAD) of 1 degree Celsius.

Here is a list of the high temperatures, in degrees Celsius, for the second week;

5, 9, 7, 0, 3, 5, 5

(A) The mean temperature for the second week is given by;

Mean =
`\frac{\text{Sum of all data values}}{\text{Total number of values}}`

=
`(5+9+7+0+3+5+5)/(7)`

=
`(34)/(7)` = 4.86 degrees Celsius

(B) Mean absolute deviation for the second week is calculated as;

Firstly, subtracting mean from each data value;

| 5 - 4.86 | = 0.14

| 9 - 4.86 | = 4.14

| 7 - 4.86 | = 2.14

| 0 - 4.86 | = 4.86

| 3 - 4.86 | = 1.86

| 5 - 4.86 | = 0.14

| 5 - 4.86 | = 0.14

Now, we will find the mean of these deviations;

Mean absolute deviation =
`(0.14+4.14 + 2.14+4.86+1.86+0.14+0.14)/(7)`

=
`(13.42)/(7)` = 1.92 degrees Celsius

(C) The second week had greater variability in high temperature because it's mean absolute deviation is greater than that of first week. And also the mean absolute deviation tells us that on average how far the data values are from the mean.