Mean, median and mode questions with answers

Questions

1. Question 1The mean of a list

   The masses, in kg, of 6 durians are $3.1,\ 1.9,\ 3.4,\ 2.1,\ 3.6,\ 3.3$. Find the mean.

   Show the worked solution

   1. Step 1: Add up all the valuesSum $= 3.1 + 1.9 + 3.4 + 2.1 + 3.6 + 3.3$$= 17.4$
   2. Step 2: Mean = sum ÷ number of valuesMean $= \dfrac{17.4}{6}$$= 2.9$ kg
2. Question 2The median of a list

   The marks scored by 7 pupils in a quiz are $5,\ 12,\ 18,\ 7,\ 5,\ 19,\ 5$. Find the median.

   Show the worked solution

   1. Step 1: Arrange the values in order, smallest first$5,\ 5,\ 5,\ 7,\ 12,\ 18,\ 19$
   2. Step 2: Find the position of the median, then the medianThe middle value is the $\dfrac{7 + 1}{2} = 4$th value.Median $= 7$
3. Question 3The mode of a list

   The numbers of minutes 10 pupils took to travel to school are $9,\ 14,\ 5,\ 7,\ 6,\ 6,\ 11,\ 7,\ 11,\ 6$. Find the mode.

   Show the worked solution

   1. Step 1: Count how many times each value appears$6$ appears $3$ times, more than any other value.
   2. Step 2: The mode is the value that appears most oftenMode $= 6$ min
4. Question 4A missing value from the mean

   The numbers of goals scored by a team in 6 matches are $2,\ x,\ 0,\ 4,\ 6,\ 1$. The mean is $3$. Find $x$.

   Show the worked solution

   1. Step 1: Total of all the values = mean × number of valuesTotal $= 3 \times 6$$= 18$
   2. Step 2: Add up the values you knowSum of the others $= 2 + 0 + 4 + 6 + 1$$= 13$
   3. Step 3: The missing value is what is left of the total$x = 18 - 13$$= 5$
5. Question 5Adding a value: the new mean

   The mean mass of 7 parcels is $11$ kg. When the mass of one more parcel is included, the mean mass of the 8 parcels becomes $10$ kg. Find the mass of this parcel.

   Show the worked solution

   1. Step 1: The total of the values given first: mean × number of themTotal $= 11 \times 7$$= 77$
   2. Step 2: The total for all 8: new mean × new numberNew total $= 10 \times 8$$= 80$
   3. Step 3: The added value is the difference in the totalsAdded value $= 80 - 77$$= 3$ kg
6. Question 6Every value changed by the same amount

   The mean, median, mode and range of the ages, in years, of the 13 members of a choir are $43.7$ years, $47$ years, $42$ years and $36$ years. $5$ years later, all the members are still in the choir and the statistics are worked out again. Find the new mean.

   Show the worked solution

   1. Step 1: Adding the same amount to every value moves every average by that amount; the spread (range) does not changeEvery value goes up by $5$, so the mean goes up by $5$.
   2. Step 2: The new meanNew mean $= 43.7 + 5$$= 48.7$ years
7. Question 7Which average? Will the mean go up or down?

   The list shows the sizes of the school T-shirts ordered by 13 pupils: $\text{L},\ \text{L},\ \text{S},\ \text{L},\ \text{L},\ \text{M},\ \text{M},\ \text{L},\ \text{XL},\ \text{M},\ \text{L},\ \text{S},\ \text{L}$. The bookshop wants to know which size to keep the most of. Which average, the mean, the median or the mode, is the most suitable to use?

   Show the worked solution

   1. Step 1: Can every average be worked out?The sizes are not numbers, so the mean cannot be calculated; only the most common size can be found.
   2. Step 2: The mode is the most common valueThe mode is the most suitable: it is the most common size.
8. Question 8Two values added: from the new mean and mode

   The numbers of books read last month by 5 pupils are $2,\ 5,\ 8,\ 13,\ 14$. Two more pupils are asked. For all $7$ pupils, the mode is $2$ and the mean is $8$. Find the two new values.

   Show the worked solution

   1. Step 1: The total of all 7 values: mean × number of valuesTotal $= 8 \times 7$$= 56$
   2. Step 2: The two new values add up to what is leftSum of the 5 values given $= 2 + 5 + 8 + 13 + 14$$= 42$The two new values add up to $56 - 42 = 14$.
   3. Step 3: The mode must appear more than once: one new value is the modeOne new value is $2$.The other is $14 - 2 = 12$.
9. Question 9Five numbers from their mode, median, mean and range

   The numbers of goals scored by a team in five matches have a mode of $3$, a median of $5$, a mean of $6$ and a range of $9$. Find the five numbers, from the smallest to the largest.

   Show the worked solution

   1. Step 1: The median is the middle (3rd) numberThe 3rd number is the median, $5$.
   2. Step 2: The mode must appear at least twice; it is less than the median, so it is the 1st and 2nd numbersThe 1st and 2nd numbers are both $3$.
   3. Step 3: Range = largest − smallest5th number $= 3 + 9$$= 12$
   4. Step 4: The total of the five numbers is mean × 5: the 4th number makes it upTotal $= 6 \times 5$$= 30$4th number $= 30 - 3 - 3 - 5 - 12$$= 7$The numbers are $3,\ 3,\ 5,\ 7,\ 12$ (only the mode, $3$, appears twice).