Mystery Boxes: Breakout Rooms

The boxes below contain 14 numbers, listed in order, of which 6 have been removed. Your job is

to use the clues given to determine the missing numbers. Write your answers in the boxes.

• The mean is 26

• The median is 22

The interquartile range is 20

The mode is 18

. The range is 60

1

4

15

18

29

30

32

58

Answer:

[tex]\begin{array}{ccccccccccccccc}{1} & {3} & {4} & {12} & {15} & {18}& {18 } & {26} & {29} & {30} & {32} & {57} & {58} & {61} \\ \end{array}[/tex]

Step-by-step explanation:

Given

[tex]\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {[ \ ]} \\ \end{array}[/tex]

Required

Fill in the box

From the question, the range is:

[tex]Range = 60[/tex]

Range is calculated as:

[tex]Range = Highest - Least[/tex]

From the box, we have:

[tex]Least = 1[/tex]

So:

[tex]60 = Highest - 1[/tex]

[tex]Highest = 60 +1[/tex]

[tex]Highest = 61[/tex]

The box, becomes:

[tex]\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}[/tex]

From the question:

[tex]IQR = 20[/tex] --- interquartile range

This is calculated as:

[tex]IQR = Q_3 - Q_1[/tex]

[tex]Q_3[/tex] is the median of the upper half while [tex]Q_1[/tex] is the median of the lower half.

So, we need to split the given boxes into two equal halves (7 each)

Lower half:

[tex]\begin{array}{ccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } \\ \end{array}[/tex]

Upper half

[tex]\begin{array}{ccccccc}{[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}[/tex]

The quartile is calculated by calculating the median for each of the above halves is calculated as:

[tex]Median = \frac{N + 1}{2}th[/tex]

Where N = 7

So, we have:

[tex]Median = \frac{7 + 1}{2}th = \frac{8}{2}th = 4th[/tex]

So,

[tex]Q_3[/tex] = 4th item of the upper halves

[tex]Q_1[/tex]= 4th item of the lower halves

From the upper halves

[tex]\begin{array}{ccccccc}{[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}[/tex]

We have:

[tex]Q_3 = 32[/tex]

[tex]Q_1[/tex] can not be determined from the lower halves because the 4th item is missing.

So, we make use of:

[tex]IQR = Q_3 - Q_1[/tex]

Where [tex]Q_3 = 32[/tex] and [tex]IQR = 20[/tex]

So:

[tex]20 = 32 - Q_1[/tex]

[tex]Q_1 = 32 - 20[/tex]

[tex]Q_1 = 12[/tex]

So, the lower half becomes:

Lower half:

[tex]\begin{array}{ccccccc}{1} & {[ \ ]} & {4} & {12 } & {15} & {18}& {[ \ ] } \\ \end{array}[/tex]

From this, the updated values of the box is:

[tex]\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {12} & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}[/tex]

From the question, the median is:

[tex]Median = 22[/tex] and [tex]N = 14[/tex]

To calculate the median, we make use of:

[tex]Median = \frac{N + 1}{2}th[/tex]

[tex]Median = \frac{14 + 1}{2}th[/tex]

[tex]Median = \frac{15}{2}th[/tex]

[tex]Median = 7.5th[/tex]

This means that, the median is the average of the 7th and 8th items.

The 7th and 8th items are blanks.

However, from the question; the mode is:

[tex]Mode = 18[/tex]

Since the values of the box are in increasing order and the average of 18 and 18 do not equal 22 (i.e. the median), then the 7th item is:

[tex]7th = 18[/tex]

The 8th item is calculated as thus:

[tex]Median = \frac{1}{2}(7th + 8th)[/tex]

[tex]22= \frac{1}{2}(18 + 8th)[/tex]

Multiply through by 2

[tex]44 = 18 + 8th[/tex]

[tex]8th = 44 - 18[/tex]

[tex]8th = 26[/tex]

The updated values of the box is:

[tex]\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {12} & {15} & {18}& {18 } & {26} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}[/tex]

From the question.

[tex]Mean = 26[/tex]

Mean is calculated as:

[tex]Mean = \frac{\sum x}{n}[/tex]

So, we have:

[tex]26= \frac{1 + 2nd + 4 + 12 + 15 + 18 + 18 + 26 + 29 + 30 + 32 + 12th + 58 + 61}{14}[/tex]

Collect like terms

[tex]26= \frac{ 2nd + 12th+1 + 4 + 12 + 15 + 18 + 18 + 26 + 29 + 30 + 32 + 58 + 61}{14}[/tex]

[tex]26= \frac{ 2nd + 12th+304}{14}[/tex]

Multiply through by 14

[tex]14 * 26= 2nd + 12th+304[/tex]

[tex]364= 2nd + 12th+304[/tex]

This gives:

[tex]2nd + 12th = 364 - 304[/tex]

[tex]2nd + 12th = 60[/tex]

From the updated box,

[tex]\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {12} & {15} & {18}& {18 } & {26} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}[/tex]

We know that:

The 2nd value can only be either 2 or 3

The 12th value can take any of the range 33 to 57

Of these values, the only possible values of 2nd and 12th that give a sum of 60 are:

[tex]2nd = 3[/tex]

[tex]12th = 57[/tex]

i.e.

[tex]2nd + 12th = 60[/tex]

[tex]3 + 57 = 60[/tex]

So, the complete box is:

[tex]\begin{array}{ccccccccccccccc}{1} & {3} & {4} & {12} & {15} & {18}& {18 } & {26} & {29} & {30} & {32} & {57} & {58} & {61} \\ \end{array}[/tex]

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