1. **State the problem:** We have a table showing the number of guests in different age groups at a wedding. We need to (a) draw a histogram, (b) find the median age group, (c) find the smallest possible number of guests under 50 years, and (d) estimate the mean age using mid-interval values. 2. **Draw the histogram:** - The horizontal axis represents age groups: 0-20, 20-40, 40-60, 60-80, 80-100. - The vertical axis represents the number of guests. - Draw bars for each age group with heights equal to the number of guests: 9, 58, 42, 27, and 4 respectively. 3. **Find the median age group:** - Total guests = $9 + 58 + 42 + 27 + 4 = 140$. - Median position = $\frac{140 + 1}{2} = 70.5$th guest. - Cumulative guests by age group: - 0-20: 9 - 20-40: 9 + 58 = 67 - 40-60: 67 + 42 = 109 - The 70.5th guest lies in the 40-60 age group because 67 < 70.5 \leq 109. 4. **Smallest possible number of guests under 50 years:** - Age groups under 50 years are 0-20 and part of 40-60 (since 40-60 covers 40 to 60). - Guests in 0-20: 9 - Guests in 20-40: 58 (all under 40, so under 50) - For 40-60 group (42 guests), only those aged 40 to 50 count. - The 40-60 group spans 20 years; 40-50 is 10 years, half of the group. - Assuming uniform distribution, half of 42 = 21 guests are under 50. - Total under 50 = 9 + 58 + 21 = 88 guests. 5. **Estimate the mean age:** - Use mid-interval values: 10, 30, 50, 70, 90. - Multiply mid-interval by number of guests: - $10 \times 9 = 90$ - $30 \times 58 = 1740$ - $50 \times 42 = 2100$ - $70 \times 27 = 1890$ - $90 \times 4 = 360$ - Sum of products = $90 + 1740 + 2100 + 1890 + 360 = 6180$ - Mean age = $\frac{6180}{140} = 44.142857...$ - Rounded to 1 decimal place: $44.1$ years. **Final answers:** - (b) Median age group: 40-60 years. - (c) Smallest possible number under 50 years: 88 guests. - (d) Estimated mean age: 44.1 years.