Grade 12 - More exercises : answers.

Mean, median, mode and range.

Number Mean Median Mode Range
1.1 20,909 20 19 12
1.2 14,5 17 17 11
1.3 55,333 59 60 25
1.4 50,5 51 50 and 51 31
1.5 25 25 no mode 17
1.6 18,333 19 19 10
1.7 9,333 11 15 21
1.8 31,833 33 37 7
2.1
Mean = 34   ;   median = 34   ;   mode = 34   and range = 6
The data values are not too spread out and therefore all three the measures, the mean, median
and mode, describe the data well.
2.2
Mean = 9,889   ;   median = 7   ;   mode = 7   and range = 25
The outlier 31 affects the mean. The mean is thus not a reliable description of the data.
The median and mode are better descriptions.
2.3
Mean = 12,667   ;   median = 15   ;   mode = 15   and range = 13
The outlier 3 affects the mean. The mean is therefore not a reliable description of the data.
The median and mode are better descriptions.
2.4
Mean = 19,667   ;   median = 20   ;   mode = 20   and range = 6
The mean is not too bad   —   it is influenced by the values that are slightly slanted to the left.
The median and mode are good descriptions.
2.5
Mean = 28   ;   median = 23   ;   mode = 22   and range = 41
The outlier 62 has a great influence on the mean. The mean is therefore not a reliable description
of the data. The mode is slightly too small. The median is the best description.
2.6
Mean = 52,5   ;   median = 52,5   ;   mode = no modus   and range = 79
The outliers 13 and 92 "balance" one another so that the mean is a good description of the data.
The median is also a good description of the data. The mode does not exist and can therefore
not be used.
3.
Number 3.1 3.2 3.3 3.4
median 23,5 63 14 40,5
1st quartile, Q1 16 53 9 35
2nd quartile, Q2 23,5 63 14 40,5
3rd quartile, Q3 35 75 31 44
inter quartile width 19 22 22 9
20th percentile, P20 15 51 7 34
25th percentile, P25 16 53 9 35
75th percentile, P75 35 75 31 44
80th percentile, P80 36 92 34 48
boundaries for middle 50% 16 to 35 53 to 75 9 to 31 35 to 44
maximum for bottom 25% 16 53 9 35
minimum for top 20% 36 92 34 48
4.
Perhaps the following can help: There are 5 different values and therefore we can write it:
a   ;   b   ;   c   ;   d   ;   e
We also know that the greatest value is 28, that the median is 24 and that the range is 10
The median is the value exactly in the middle and therefore c = 24 and a = 28 – 10
Therefore the values are:   a   ;   b   ;   24   ;   d   ;   28
4.1
The smallest value   =   greatest value   –   range
  =   28   –   10   =   18
The values are 18   ;   b   ;   24   ;   d   ;   28
4.2
There is no mode because the values are all different.
4.3
The values are 18   ;   b   ;   24   ;   d   ;   28
There are 2 values smaller than the median and 2 greater than the median.
4.4
One. Look at the values that we have written down. Only d is between the median, 24, and the
greatest value, 28.
4.5
Approximate sum = mean X number of values
= 23,2 X 5 = 116
5.
Again write down a set of values to assist: a   ;   b   ;   c   ;   d   ;   e   ;   f   ;   g
Replace with known values: 14   ;   b   ;   c   ;   17   ;   e   ;   f   ;   g
5.1
Greatest value = smallest value + range = 14 + 6 = 20
5.2
The values are now 14   ;   b   ;   c   ;   17   ;   e   ;   f   ;   20
Three values are smaller than 17 and 3 are greater than 17. There are 7 values and the median
is the value exactly in the middle, viz. 17.
5.3
Sum = 17,143 x 7 = 120
5.4
There are 3 values smaller than 17 and 14 is the smallest value. Therefore there are 2 values
between 14 and 17 .
5.5
There are 3 values greater than 17 and 20 is the greatest value. Therefore are 2 values
between 17 and 20. The mode is 19 and it is the value with the highest frequency.
There are 2 unknown values and therefore they must be 19 and 19.
6.
Again write down a set of values to assist:  7   ;   b   ;   c   ;   d   ;   17   ;   f   ;   g   ;   h   ;   i
6.1
Biggest value = 7 + 26 = 33
6.2
The values are:  7   ;   b   ;   c   ;   d   ;   17   ;   f   ;   g   ;   h   ;   33
The median is the central value and it is 17.
There are 4 values smaller than the median and 4 values greater than the median.
6.3
The mean is 18 and all the values are integers. The mean, 18, is therefore the
first value following the median, 17. The values are:
  7   ;   b   ;   c   ;   d   ;   17   ;   18   ;   g   ;   h   ;   33.
Therefore there are 3 values greater than the mean.
6.4
All the values are integers so that the median, 17, is the first number greater than 16.
The mode is 27. There are two values between 18 and 33 and therefore g = h = 27.
The numbers greater than 16 are: 17   ;   18   ;   27   ;   27   ;   33
6.5
Sum of all the values = mean X number of values
= 18 X 9   = 162
6.6
The sum of the values in 6.4 = 122
6.7
The sum of all the values smaller than the median = (sum of all the values) – (sum in 6.6)
= 40
7.
Again write down a set to assist us:
2   ;   b   ;   c   ;   d   ;   e   ;   (9)   ;   f   ;   g   ;   h   ;   i   ;   j    — (9) is the mode
7.1
The biggest value = 2 + 19 = 21
7.2
According to our set there are 5 values smaller and 5 values greater than the median, 9, because the
median is the midmost value.
7.3
Yes. The median is the midmost value and thus half, or 50%, of the values are smaller than
the median.
7.4
It depends on the value of f. If f = 10, there are 4 values greater than the mean, 10,5.
If f ≥ 11, there are 5 values greater than the mean.
7.5
The approximate sum of all the values = 10 X 10,5 = 105
7.6
The mode has a frequency of 2 and a value of 8. According to our set, there are
5 values smaller than 9. Two of these values are 8. There are 3 values smaller than 8.
Three values are smaller than the mode.
8.1
Total number of runs = 23,1 x 7 = 161,7,   i.e. 162 runs
Subtract his first 6 scores   —   he scored 31 runs in the 7th match.
8.2
No. Four scores are below 10 and 5 scores are below the mean.
8.3
Mean = 138 ÷ 7   =   19,7
8.4
The second batsman. His scores were consistently close to his average.
  
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