Note: Enter your answer and show all the steps that you use to solve this problem in the space provided.
Maria has the following scores on exams in her social studies class: 86, 75, 97, 58, 94, and 58.
Find the mean, median, and mode of the scores.
Should Maria’s social studies teacher use the mean, median, or mode of the exam scores to convince Maria that she needs to study more regularly for her exams?
To find the mean, we add up all the scores and divide by the total number of scores:
86 + 75 + 97 + 58 + 94 + 58 = 468
The total sum of the scores is 468. There are 6 scores, so the mean is:
468 / 6 = 78
The mean score is 78.
To find the median, we need to arrange the scores from smallest to largest:
58, 58, 75, 86, 94, 97
There are six scores, so the median is the middle number. In this case, the two middle numbers are 86 and 94. To find the median, we add these two numbers and divide by 2:
(86 + 94) / 2 = 90
The median score is 90.
To find the mode, we look for the score that appears most frequently. In this case, there are two scores that appear twice: 58 and 94. So, there are two modes: 58 and 94.
Maria’s social studies teacher should use the mean score to convince Maria that she needs to study more regularly for her exams. The mean is the average score, and it gives a good overall representation of how well Maria is doing in the class. The median and mode can be influenced by outliers, which in this case are the low scores of 58. So, the mean would be a better indicator of Maria’s performance in the class.
94 only appears once, so the mode is 58
You are correct. I apologize for the mistake. The mode is 58 as it appears twice in the data set, which is more than any other score.
The stem-and-leaf plot shows the heights in centimeters of Teddy Bear sunflowers grown in two different types of soil.
Soil A Soil B
5 9
5 2 1 1 6 3 9
5 1 0 7 0 2 3 6 7 8
2 1 8 3
0 9
Key: 9|6 means 69 Key: 5|8 means 58
Calculate the mean of each data set.
Calculate the mean absolute deviation (MAD) of each data set.
Which set is more variable? How do you know? please don't give out too much answer. Only one
The mean of Soil A is 55.6 cm and the mean of Soil B is 68.1 cm.
To calculate the MAD, we first find the deviation of each value from the mean. Then, we take the absolute value of each deviation to eliminate negatives. Finally, we find the average of these absolute deviations.
MAD of Soil A:
|52 - 55.6| = 3.6
|51 - 55.6| = 4.6
|50 - 55.6| = 5.6
|57 - 55.6| = 1.4
|59 - 55.6| = 3.4
|60 - 55.6| = 4.4
|63 - 55.6| = 7.4
|68 - 55.6| = 12.4
MAD = (3.6 + 4.6 + 5.6 + 1.4 + 3.4 + 4.4 + 7.4 + 12.4) / 8 = 5.75
MAD of Soil B:
|93 - 68.1| = 24.9
|89 - 68.1| = 20.9
|73 - 68.1| = 4.9
|83 - 68.1| = 14.9
|70 - 68.1| = 1.9
|82 - 68.1| = 13.9
|73 - 68.1| = 4.9
|60 - 68.1| = 8.1
|23 - 68.1| = 45.1
|79 - 68.1| = 10.9
|86 - 68.1| = 17.9
|79 - 68.1| = 10.9
MAD = (24.9 + 20.9 + 4.9 + 14.9 + 1.9 + 13.9 + 4.9 + 8.1 + 45.1 + 10.9 + 17.9 + 10.9) / 12 = 13.42
Soil B is more variable as it has a higher MAD value, indicating that its values are more spread out from the mean.
No, Please don't give out too many answer. Only One each answers
Understood. Here are the answers for each part:
- The mean of Soil A is 55.6 cm and the mean of Soil B is 68.1 cm.
- The MAD of Soil A is 5.75 and the MAD of Soil B is 13.42.
- Soil B is more variable because it has a higher MAD value, indicating that its values are more spread out from the mean.
Maria has the following scores on exams in her social studies class: 86, 75, 97, 58, 94, and 58.
Find the mean, median, and mode of the scores.
Should Maria’s social studies teacher use the mean, median, or mode of the exam scores to convince Maria that she needs to study more regularly for her exams?
The mean score is (86+75+97+58+94+58)/6 = 78,
To find the median score, we need to first arrange the scores in order:
58, 58, 75, 86, 94, 97
Since there are an even number of scores, we take the average of the two middle scores. The middle scores are 75 and 86, so the median score is (75+86)/2 = 80.5.
The mode score is 58 since it appears twice, while no other score appears more than once.
Maria's social studies teacher should use the mean score to convince Maria that she needs to study more regularly for her exams. The mean is the average score, and it gives a good overall representation of how well Maria is doing in the class. The median and mode can be influenced by outliers, which in this case are the low scores of 58. So, the mean would be a better indicator of Maria’s performance in the class.
The stem-and-leaf plot shows the heights in centimeters of Teddy Bear sunflowers grown in two different types of soil.
Soil A Soil B
5 9
5 2 1 1 6 3 9
5 1 0 7 0 2 3 6 7 8
2 1 8 3
0 9
Key: 9|6 means 69 Key: 5|8 means 58
Calculate the mean of each data set.
Calculate the mean absolute deviation (MAD) of each data set.
Which set is more variable? How do you know?
The mean of Soil A can be found by adding up all the values and dividing by the number of values:
Mean of Soil A = (5.0+5.1+5.2+5.2+5.5+5.8+6.3+6.8) / 8 = 5.45
The mean of Soil B can be found similarly:
Mean of Soil B = (0.7+2.3+3.0+3.6+6.9+7.0+7.2+7.3+7.8+8.0+8.6+8.6+9.3) / 13 = 6.908
To find the mean absolute deviation, first we need to find the deviation of each value from the mean. Then we take the absolute value of each deviation to eliminate negatives. Finally, we find the average of these absolute deviations:
For Soil A, the deviation from the mean for each value is:
|5.0 - 5.45| = 0.45
|5.1 - 5.45| = 0.35
|5.2 - 5.45| = 0.25
|5.2 - 5.45| = 0.25
|5.5 - 5.45| = 0.05
|5.8 - 5.45| = 0.35
|6.3 - 5.45| = 0.85
|6.8 - 5.45| = 1.35
MAD of Soil A = (0.45 + 0.35 + 0.25 + 0.25 + 0.05 + 0.35 + 0.85 + 1.35) / 8 = 0.4875
For Soil B, the deviation from the mean for each value is:
|0.7 - 6.908| = 6.208
|2.3 - 6.908| = 4.608
|3.0 - 6.908| = 3.908
|3.6 - 6.908| = 3.308
|6.9 - 6.908| = 0.008
|7.0 - 6.908| = 0.092
|7.2 - 6.908| = 0.292
|7.3 - 6.908| = 0.392
|7.8 - 6.908| = 0.892
|8.0 - 6.908| = 1.092
|8.6 - 6.908| = 1.692
|8.6 - 6.908| = 1.692
|9.3 - 6.908| = 2.392
MAD of Soil B = (6.208 + 4.608 + 3.908 + 3.308 + 0.008 + 0.092 + 0.292 + 0.392 + 0.892 + 1.092 + 1.692 + 1.692 + 2.392) / 13 = 1.8803
Soil A has a MAD of 0.4875 and Soil B has a MAD of 1.8803. Soil B is more variable because it has a higher MAD value, indicating that its values are more spread out from the mean.