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Mean, Median, Mode, and Range: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 The instrument log

An instrument log contains $8$ readings whose sum is $-12$ . What is the mean of the readings?
1. Hint 1
  The mean is the amount each reading would have if the total were shared equally.
2. Hint 2
  Divide the given sum by the number of readings, keeping the sign of the sum.
Answer

$-1.5$ , or $-\tfrac32$ .

Full solution

The sum and the number of readings are already known, so divide the sum by $8$ .
$\text{mean}=\frac{-12}{8}$ $\text{mean}=-1.5$

Replacing all eight readings by $-1.5$ gives the original total:
$8(-1.5)=-12$

Answer

$-1.5$ , or $-\tfrac32$ .

Key idea

The sum and the number of values are enough to determine the mean.
Problem 1: Mark complete
Problem 2 The covered reading

Six readings are $14$ , $22$ , $16$ , $18$ , $20$ , and one covered value. The covered value is less than $14$ . Find the median of the six readings.
1. Hint 1
  Locate the middle positions after putting every reading in order.
2. Hint 2
  The covered reading belongs before all five visible readings, so it still occupies one position.
3. Hint 3
  For six readings, average the third and fourth readings in order.
Answer

$17$ .

Full solution

The covered reading is the smallest.
The sorted order is the covered reading, $14$ , $16$ , $18$ , $20$ , $22$ .

The middle two readings are $16$ and $18$ .
Their mean gives the median.
$\text{median}=\frac{16+18}{2}$ $\text{median}=17$

There are three readings below $17$ and three above it, regardless of the covered value.

Answer

$17$ .

Key idea

An unknown extreme value need not prevent you from locating the middle of an ordered list.
Problem 2: Mark complete
Problem 3 The archive entry

An archive entry gives the smallest value in a data set as $-7$ and the range as $19$ . What is the largest value?
1. Hint 1
  The range measures the distance from the smallest value to the largest.
2. Hint 2
  Start at the smallest value and add the full distance to the largest.
Answer

$12$ .

Full solution

Add the range to the smallest value to reach the largest value.
$-7+19=12$

Check by subtracting the smallest from the largest.
$12-(-7)=19$

Answer

$12$ .

Key idea

The largest value equals the smallest value plus the range.
Problem 3: Mark complete
Problem 4 The water jars

Five jars contain $18$ , $16$ , $20$ , $16$ , and $15$ milliliters of water. Transfer $2$ milliliters from the third jar to the second jar without spilling any water. Find the mean, median, mode, and range of the five amounts after the transfer.
1. Hint 1
  A transfer moves water between jars without changing the total, but it changes the amounts in two of them.
2. Hint 2
  Record the five amounts that remain after the transfer before finding any summary.
3. Hint 3
  Sort the final amounts for the median, count repeats for the mode, and compare the extremes for the range.
Answer

Mean $17$ milliliters; median $18$ milliliters; mode $18$ milliliters; range $3$ milliliters.

Full solution

The third jar loses $2$ milliliters and the second jar gains $2$ milliliters.
$20-2=18$ $16+2=18$
The final amounts are $18$ , $18$ , $18$ , $16$ , and $15$ milliliters.

Add the final amounts and divide by the five jars.
$18+18+18+16+15=85$ $\text{mean}=\frac{85}{5}$ $\text{mean}=17$
The mean is $17$ milliliters.

The sorted amounts are $15$ , $16$ , $18$ , $18$ , $18$ milliliters.
The third value is $18$ , so the median is $18$ milliliters.
The value $18$ occurs three times and each other value once, so the mode is also $18$ milliliters.
The range is
$18-15=3$
or $3$ milliliters.

The original amounts sort to $15$ , $16$ , $16$ , $18$ , $20$ milliliters and add to $85$ milliliters.
The final total of $85$ milliliters matches the original total, as a transfer without spilling requires.
Before the transfer the median and mode were both $16$ milliliters and the range was $5$ milliliters, so only the mean stayed the same.

Answer

Mean $17$ milliliters; median $18$ milliliters; mode $18$ milliliters; range $3$ milliliters.

Key idea

Update the affected entries before finding summaries of a changed data set.
Problem 4: Mark complete
Problem 5 The decal batch

A print shop has a batch of seven decals: comet, leaf, comet, wave, leaf, leaf, comet. It then prints one more leaf decal for the batch. Which of the mean, median, mode, and range can be found for the seven decals, and which for all eight in the finished batch? Give the value of each one that can be found.
1. Hint 1
  Ask what each of the four summaries needs from the data before it can be found.
2. Hint 2
  Decal designs have no numerical values and no least-to-greatest order, but they can still be counted.
3. Hint 3
  Count each design before and after the new decal, and remember that a tie for the highest count, with another design counted less often, gives more than one mode.
Answer

Only the mode can be found. The seven decals have two modes, comet and leaf; the eight decals in the finished batch have one mode, leaf.

Full solution

The mean and range need numerical values, and the median needs a least-to-greatest order.
Decal designs have neither.
The mode compares how often each design occurs, so it is the only one of the four that can summarize these names.

Among the seven decals there are three comets, three leaves, and one wave.
Comet and leaf tie for the highest tally, and the wave has a lower tally, so the seven decals have two modes, comet and leaf.

Printing one more leaf gives
$3+1=4$
leaves.
The counts are now four leaves, three comets, and one wave, so leaf alone is the mode of the finished batch.

The final counts account for the eight decals.
$4+3+1=8$

Answer

Only the mode can be found. The seven decals have two modes, comet and leaf; the eight decals in the finished batch have one mode, leaf.

Key idea

The mode can summarize names by comparing how often they occur, and a tie for the highest count, while some other design is counted less often, gives more than one mode.
Problem 5: Mark complete
Problem 6 The delivery report

A delivery report covers $36$ deliveries. Their times have a mean of $28$ minutes and a median of $12$ minutes. A few deliveries included unusually long waits. A manager needs a typical delivery time and the total time for all $36$ deliveries. What should the two reported times be?
1. Hint 1
  The two requested times serve different purposes: one describes a typical delivery, and the other accounts for every minute.
2. Hint 2
  For the typical time, choose the center that resists the unusually long waits.
3. Hint 3
  Recover a total from the number of values and their mean.
Answer

Typical time: $12$ minutes, using the median; total time: $1008$ minutes.

Full solution

The mean of $28$ minutes is more than twice the median of $12$ minutes, a sign that the few unusually long waits pull the mean above the usual delivery times.
The median depends on the middle position, so the reported median of $12$ minutes is the better typical time here.

The total must include the long waits as well as every other delivery.
The mean accounts for that full total.
Multiply it by the number of deliveries.
$36\times28=1008$
The total is $1008$ minutes.

Check the total against the reported mean.
$1008\div36=28$

Answer

Typical time: $12$ minutes, using the median; total time: $1008$ minutes.

Key idea

The median can describe a typical value while the mean still determines the total.
Problem 6: Mark complete
Problem 7 The two filling stations

Station A fills $3$ containers with a mean amount of $6$ liters. Station B fills $5$ containers with a mean amount of $10$ liters. What is the mean amount in all $8$ containers together?
1. Hint 1
  Combining the containers means combining their total amounts before sharing again.
2. Hint 2
  Find the total at each station from its count and mean.
3. Hint 3
  Add the two totals and divide by the combined number of containers.
Answer

$8.5$ liters, or $\tfrac{17}{2}$ liters.

Full solution

The mean times the number of containers gives each station's total.
$3\times6=18$ $5\times10=50$
Station A contributes $18$ liters and Station B contributes $50$ liters.

Together there are $8$ containers holding
$18+50=68$
liters.
Their mean amount is
$68\div8=8.5$
or $8.5$ liters.

As a check, eight containers each holding the combined mean would contain the same total.
$8\times8.5=68$

Answer

$8.5$ liters, or $\tfrac{17}{2}$ liters.

Key idea

To combine group means, recover each group total and divide their sum by the combined count.
Problem 7: Mark complete
Problem 8 Nia's claim

A list has four different whole numbers, three below $10$ and one above $10$ . Nia says its mean could still be $10$ . Is Nia right? If she is, give one such list and show its deviations from $10$ ; if she is not, explain why no such list exists.
1. Hint 1
  Think of the mean as a balance point and ask what must balance on each side of it.
2. Hint 2
  Choose three different whole numbers below $10$ and add their distances below $10$ .
3. Hint 3
  Place the remaining number that total distance above $10$ .
Answer

Yes. One example is $6$ , $8$ , $9$ , $17$ , with deviations $-4$ , $-2$ , $-1$ , $7$ .

Full solution

Choose $6$ , $8$ , and $9$ below $10$ .
Their distances below $10$ add to
$4+2+1=7$
so choose $17$ , which is $7$ above $10$ .

The deviations are $-4$ , $-2$ , $-1$ , and $7$ .
They balance:
$-4-2-1+7=0$
This makes $10$ the balance point even though three values lie below it.

Check by finding the mean directly.
$6+8+9+17=40$ $40\div4=10$
Thus Nia is right.
Other lists that meet the conditions are also valid.

Answer

Yes. One example is $6$ , $8$ , $9$ , $17$ , with deviations $-4$ , $-2$ , $-1$ , $7$ .

Key idea

More values may lie on one side of the mean when their total deviations balance those on the other side.
Problem 8: Mark complete
Problem 9 Sam's copied list

A data list has exactly two modes. Sam copies the entire list once and attaches the copy to the original. He says the resulting list has four modes. Is Sam right? Explain.
1. Hint 1
  A mode is a value with a highest tally, not a single appearance in the list.
2. Hint 2
  Determine how attaching one full copy changes every tally.
Answer

No; the same two values remain the modes.

Full solution

Attaching one full copy doubles the tally of every value.
The two values tied for the highest tally still tie, and neither is replaced by a new value.

To check that a lower tally stays lower, let $a$ be the original highest tally and $b$ any lower tally.
Multiplying both by the same positive number preserves their order.
$2a>2b$
No other value catches up, so there are still exactly two modes.

Answer

No; the same two values remain the modes.

Key idea

Copying an entire list the same number of times preserves which values have the highest tallies.
Problem 9: Mark complete
Problem 10 Lee's claim

Lee says two lists of three whole numbers that both have range $6$ must have the same mean. Is Lee right? If he is, explain why; if he is not, give two such lists and their means.
1. Hint 1
  The range controls the gap between the smallest and largest entries.
2. Hint 2
  Choose a middle entry between two numbers $6$ apart.
3. Hint 3
  For the other list, move all three entries by the same amount and compare the new extremes.
Answer

No. One example: $0$ , $3$ , $6$ , mean $3$ ; $10$ , $13$ , $16$ , mean $13$ . Both ranges are $6$ , but the means differ.

Full solution

One possible first list is $0$ , $3$ , $6$ .
Its range and mean are
$6-0=6$ $\frac{0+3+6}{3}=3$

Choose $10$ , $13$ , $16$ for the other list.
Its range and mean are
$16-10=6$ $\frac{10+13+16}{3}=13$

Both lists have range $6$ , but their means differ, so Lee is not right.
The range describes the distance between extremes, so it does not determine where the center lies.
Any two lists of three whole numbers with range $6$ and different means are acceptable.

Answer

No. One example: $0$ , $3$ , $6$ , mean $3$ ; $10$ , $13$ , $16$ , mean $13$ . Both ranges are $6$ , but the means differ.

Key idea

Equal ranges can describe data sets whose centers are far apart.
Problem 10: Mark complete

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