From the image provided, the task involves solving problems related to mean, mode, median, and range using a set of cards numbered 14 to 19. Let's break down the problem and clarify the steps:

1. Problem Explanation

You need to create sets of six numbers from the cards provided that satisfy the given conditions regarding their mean, mode, median, and sometimes all of them together. Each card can be used only once in a set.

a) Mean = 16, Mode = 16, Median = 16

• The mean is calculated as the sum of numbers divided by 6 (the total number of cards in the set).
• The mode is the most frequently occurring number in the set.
• The median is the middle value when numbers are arranged in order.

b) Mean = 17, Mode = 18, Median = 17.5

• Similar calculations but with slightly different conditions for mode and median.

c) Mean = 16, Modes = 15 & 17 (bimodal), Median = 16

• Bimodal mode means there are two numbers that appear most frequently in the set.

Solutions Provided in the Image

a) Mean = 16, Mode = 16, Median = 16

The numbers chosen are: 14, 15, 16, 16, 17, 18

• The total is $14 + 15 + 16 + 16 + 17 + 18 = 96$, and $96 \div 6 = 16$ (mean is correct).
• Mode is 16 because it appears twice.
• Median is 16, as the middle values are 16 and 16 when arranged.

b) Mean = 17, Mode = 18, Median = 17.5

The numbers chosen are: 15, 16, 17, 18, 18, 18

• The total is $15 + 16 + 17 + 18 + 18 + 18 = 102$, and $102 \div 6 = 17$ (mean is correct).
• Mode is 18 because it appears three times.
• Median is $\frac{17 + 18}{2} = 17.5$.

c) Mean = 16, Modes = 15 & 17 (bimodal), Median = 16

The numbers chosen are: 14, 15, 15, 17, 17, 16

• The total is $14 + 15 + 15 + 16 + 17 + 17 = 96$, and $96 \div 6 = 16$ (mean is correct).
• Modes are 15 and 17, as they both appear twice.
• Median is 16, as the middle values are 15 and 17, and $\frac{15 + 17}{2} = 16$.

Challenge: Missing Numbers

The challenge at the bottom asks to find the missing number in different sets where the mean is already calculated. Let's solve 7a and 7b:

7a) Mean = 6, Numbers: $7, 9, 6, 4, 2, \_$

• Total sum = $6 \times 6 = 36$ (because the mean is 6 for 6 numbers).
• Sum of given numbers = $7 + 9 + 6 + 4 + 2 = 28$.
• Missing number = $36 - 28 = 8$.
  Answer: 8

7b) Mean = 10, Numbers: $10, 12, 8, 6, 14, \_$

• Total sum = $10 \times 6 = 60$.
• Sum of given numbers = $10 + 12 + 8 + 6 + 14 = 50$.
• Missing number = $60 - 50 = 10$.
  Answer: 10

Do you want further clarifications or explanations?

Additional Questions

1. Can you explain the steps for calculating the mean in detail again?
2. What happens if the mode is not clearly defined?
3. How can you identify a bimodal distribution from a set of data?
4. How is the range calculated, and why isn’t it used in these examples?
5. Can we create a new set of numbers for one of the conditions (e.g., 7a)?

Tip:

When solving problems involving mean, mode, and median, always check that all conditions are satisfied, and use clear organization to avoid missing any details!