Use this data for #1,2,3

Student Group A: 56, 14, 26, 28, 29, 45, 32, 80, 3, 10

Student Group B: 27, 24, 18, 19, 5, 16, 22, 10, 6, 20


Find mean of group A and B (round to tenth) and THEN find their difference
Find range of each student group and tell which group is more inconsistent
Find the median of each student group, and find their difference in medians
A large company wants to find the favorite animal of their employees…would surveying every 5th woman who uses the copier, be a random, representative sample? Explain your answer.
50 students are randomly selected and 4 own cats. If there are 800 students, use a proportion to estimate how many of them own cats?

1 answer

Let's break down each of the problems step by step.

1. Find the Mean of Group A and Group B, Then Their Difference

Group A: 56, 14, 26, 28, 29, 45, 32, 80, 3, 10
Group B: 27, 24, 18, 19, 5, 16, 22, 10, 6, 20

Mean of Group A: \[ \text{Mean} = \frac{\text{Sum of values}}{\text{Number of values}} = \frac{56 + 14 + 26 + 28 + 29 + 45 + 32 + 80 + 3 + 10}{10} \] \[ \text{Mean} = \frac{ 56 + 14 + 26 + 28 + 29 + 45 + 32 + 80 + 3 + 10 }{10} = \frac{ 56 + 14 + 26 + 28 + 29 + 45 + 32 + 80 + 3 + 10 }{10} = \frac{ 3 - 3 + 80 - 80 + 20 - 20 }{10} = \frac{ 54 }{10} = 32.0 \]

Mean of Group B: \[ \text{Mean} = \frac{27 + 24 + 18 + 19 + 5 + 16 + 22 + 10 + 6 + 20}{10} = \frac{ 27 + 24 + 18 + 19 + 5 + 16 + 22 + 10 + 6 + 20 }{10} \] \[ = \frac{ 27 + 24 + 18 + 19 + 5 + 16 + 22 + 10 + 6 + 20 }{10} = \frac{ 27 + 24 + 18 + 19 + 5 + 16 + 22 + 10 + 6 + 20 }{10} = 11.7 (approx) \]

Difference between means: \[ \text{Difference} = \text{Mean of Group A} - \text{Mean of Group B} = 32.0 - 17.7 = 14.3 \]

2. Find the Range of Each Student Group and Determine Which is More Inconsistent

Range of Group A: \[ \text{Range} = \text{Maximum} - \text{Minimum} = 80 - 3 = 77 \]

Range of Group B: \[ \text{Range} = \text{Maximum} - \text{Minimum} = 27 - 5 = 22 \]

Inconsistency Assessment: Group A has a range of 77, while Group B has a range of 22. Since a larger range indicates more variability in values, Group A is more inconsistent than Group B.

3. Find the Median of Each Student Group, Then Their Difference

Median of Group A: Sorted Group A: 3, 10, 14, 26, 28, 29, 32, 45, 56, 80
Median (the average of the 5th and 6th values in sorted list): \[ \text{Median} = \frac{28 + 29}{2} = \frac{57}{2} = 28.5 \]

Median of Group B: Sorted Group B: 5, 6, 10, 16, 18, 19, 20, 22, 24, 27
Median (the average of the 5th and 6th values in sorted list): \[ \text{Median} = \frac{18 + 19}{2} = \frac{37}{2} = 18.5 \]

Difference between Medians: \[ \text{Difference} = \text{Median of Group A} - \text{Median of Group B} = 28.5 - 18.5 = 10.0 \]

4. Surveying Every 5th Woman using the Copier

Surveying every 5th woman who uses the copier is not a random, representative sample. This method may introduce bias as it is likely to exclude women who do not use the copier or those who use it at different times, potentially skewing the results of the survey. A truly random sample should select individuals without any systematic pattern.

5. Estimate the Number of Students Who Own Cats

Using the proportion of sampled students who own cats:

\[ \text{Proportion} = \frac{4 \text{ (students who own cats)}}{50 \text{ (total sampled students)}} \]

To find the estimated number of students who own cats among 800 students: \[ \text{Estimated number} = \text{Proportion} \times 800 = \left( \frac{4}{50} \right) \times 800 = \frac{4 \times 800}{50} = \frac{3200}{50} = 64 \]

So the estimated number of students that own cats is 64 students.