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.