Question

question #7
mrs. cullipher and ms. moore decide to hold a friendly competition between their classes using scores from the last chapter test. the results are shown in the table.
mrs. cullipher: 74, 96, 93, 62, 83, 86, 90, 98, 93
ms. moore: 90, 91, 80, 90, 86, 76, 82, 84, 70
which statement would make the most valid argument in deciding the winner of the competition?
a mrs. culliphers class won since their interquartile range is 8 more than ms. moores class.
b mrs. culliphers class won since their standard deviation is 13 more than ms. moores class.
c mrs. culliphers class won since their median score is 4 more than ms. moores class.
d mrs. culliphers class won since their mode score is about 5 more than ms. moores class.
question #8
a choice - matrix is shown. complete the choice matrix by selecting the characteristic that best describes each distribution.
the table shows the time, in hours, george and jalecia spent studying each day last week.
georges study time (in hours): 2, 6, 3, 2, 3, 7, 4
jalecias study time (in hours): 1, 2, 3, 1, 2, 2, 3
complete the choice matrix by selecting the correct characteristic for each distribution.
jalecias study: skewed left, skewed right, no skew
georges study: skewed left, skewed right, no skew

Explanation:

Question #7

Step1: Recall measure - of - center importance

When determining a winner in a competition based on scores, measures of center like median are more relevant than measures of spread like inter - quartile range and standard deviation. The mode may not be as reliable if there are multiple modes or if the mode doesn't represent the overall performance well.

Step2: Analyze each option

• Option A: Inter - quartile range is a measure of spread, not a good indicator of overall performance to decide a winner.
• Option B: Standard deviation is a measure of spread, not ideal for deciding a winner.
• Option C: Median is a measure of center. If Mrs. Cullipher's class has a higher median score, it indicates that, on average (in a median - based sense), her class performed better. This is a valid argument for deciding a winner.
• Option D: The mode may not represent the overall class performance well.

Step1: Arrange George's data in ascending order

George's data: 2, 2, 3, 3, 4, 6, 7. The median is the middle value. Since there are 7 values, the median is the 4th value, which is 3. The mean is $\frac{2 + 2+3 + 3+4 + 6+7}{7}=\frac{27}{7}\approx3.86$. Since the mean ($3.86$) is greater than the median (3), the data is skewed right.

Step2: Arrange Jalecia's data in ascending order

Jalecia's data: 1, 1, 2, 2, 2, 3, 3. The median is the 4th value, which is 2. The mean is $\frac{1 + 1+2 + 2+2 + 3+3}{7}=\frac{14}{7} = 2$. Since the mean (2) is equal to the median (2), the data has no skew.

Answer:

C. Mrs. Cullipher's class won since their median score is 4 more than Ms. Moore's class.

Question #8