15 questions · timed · auto-graded
Solution:
The difference between the highest value and the lowest value in the data set is called Range.
Solution:
Data: 0, 2, 2, 2, -3, 5, -1, 5, 5, -3, 6, 6, 5, 6
Rearranging data in increasing order, we have
-3, -3, -1, 0, 2, 2, 5, 5, 5, 5, 6, 6, 6
Number of observations = n = 14 (even)
Now,
$\text{Median}=\frac{\Big(\frac{\text{n}}{2}\Big)^{\text{th}}\text{observation}+\Big(\frac{\text{n}+1}{2}\Big)^{\text{th}}\text{observation}}{2}$
$=\frac{7^{\text{th}}\text{observation}+8^{\text{th}}\text{observation}}{2}$
$=\frac{2+5}{2}$
$\Rightarrow\text{Median}=3.5$
$\text{k}\overline{\text{X}}$
$\frac{\overline{\text{X}}}{\text{k}}$
$\overline{\text{X}}+\text{k}$
$\overline{\text{X}}-\text{k}$
Solution:
$\text{Mean}=\overline{\text{X}}=\frac{\text{Sum of all observations}}{\text{Total number of observations}}$
$=\frac{\text{Sum of all observations}}{\text{n}}$
if each observation is multiplied by k, then
$\text{New Mean},\overline{\text{X'}}=\frac{(\text{Sum of all observations})\text{k}}{\text{n}}$
$\Rightarrow\overline{\text{X}'}=\text{k}\overline{\text{X}}$
Solution:
$\text{Mean}=\overline{\text{X}}=\frac{\text{Sum of all observations}}{\text{Total number of observations}}$
$=\frac{\text{Sum of all observations}}{\text{n}}$
Now if k is aded to each observation
$\text{New Mean},\overline{\text{X'}}=\frac{\text{Sum of all observations}+\text{nk}}{\text{n}}$
$=\frac{\text{Sum of all observations}}{\text{n}}+\text{k}$
$\Rightarrow\overline{\text{X}'}=\overline{\text{X}}+\text{k}$
Solution:
$\text{Mean}=\frac{\text{a}+\text{b}+\text{c}+\text{d}+\text{e}}{5}=28$
⇒ a + b + c + d + e = 140 ...(1)
Also, $\text{Mean}=\frac{\text{a}+\text{c}+\text{e}}{3}=24$
⇒ a+ c + e = 72 ...(2)
Subtracting equation (2) from (1), we have
b + d = 68
$\text{Mean}=\frac{\text{b}+\text{d}}{2}=\frac{68}{2}=34$
Solution:
$\text{Mean}=81=\frac{\text{Sum of seven numbers}}{7}$
⇒ Sum of seven numbers = 81 × 7 = 567
Let the discared number be x.
⇒ Sum of 6 numbers = 567 - x
Now, mean of remaining 6 numbers
$=\frac{567-\text{x}}{6}=78$⇒ 567 - x = 468
⇒ x = 99
So, discarded number is 99
Solution:
In data 5, 4, 7, 5, 8, 8, 8, 5, 7, 9, 5, 7, 9, 10, 8,
We observe that values 5 and 8 both have maximum frequency i.e. 4
Solution:
The empirical Relation between mean, median and mode is:
Mode = 3 Median - 2 mean.
Solution:
if is the mean of n observations x1, x2, x3, x4 ...xn.
then algebraic sum of deviations $=\sum\limits^\text{n}_{\text{i}=0}\Big(\text{x}_\text{i}-{\overline{\text{X}}}\Big)$
$=\sum\limits^\text{n}_{\text{i}=0}\text{x}_\text{i}-\text{n}{\overline{\text{X}}}$
$=\text{n}\bigg(\frac{\sum^\text{n}_{\text{i}=0}\text{x}_\text{i}}{\text{n}}\bigg)-\text{n}\overline{\text{X}}$
$=\text{n}\overline{\text{X}}-\text{n}\overline{\text{X}}$
$=0$
Solution:
Most Frequent value is called mode.
Solution:
Mean of first five observations $=\frac{\text{x}+\text{x}+2+\text{x}+4+\text{x}+6+\text{x}+8}{5}=11$
⇒ 5x + 20 = 55
⇒ x = 7
⇒ First three numbers are 7, 9, 11
$\text{Mean}=\frac{7+9+11}{3}=\frac{27}{3}=9$
Solution:
$\text{Mean}=\frac{7+5+13+\text{x}+9}{5}=10$
$\Rightarrow34+\text{x}=50$
$\Rightarrow\text{x}=16$
Solution:
| | Mean | Median | Mode |
| 2, 2, 2, 2, 4 | $\frac{12}{5}=2.4$ | 2 | 2 |
| 1, 3, 3, 3, 5 | $\frac{15}{5}=3$ | 3 | 3 |
| 1, 1, 2, 5, 6 | $\frac{15}{5}=3$ | 2 | 1 |
| 1, 1, 1, 2, 5 | $\frac{10}{5}=2$ | 1 | 1 |
From above table, data 1, 3, 3, 3, 5 has mean, median, mode all have same value, i.e. 3
Solution:
Median = 4
Mode = 2
$\text{Mean}=\frac{2+2+4+5+12}{5}=\frac{25}{5}=5$
Hence, (Mean = 5) > (Mode = 2)
Solution:
$\text{Mean of A}=\frac{1+2+2+3+3+3+7}{7}$
$=\frac{21}{7}=3$
$\text{Mean of B}=\frac{3+5+5+7+8+9+12}{7}$
$=\frac{49}{7}=7$
$\text{Mean of C}=\frac{2+3+4+4+4+7+11}{7}$
$=\frac{35}{7}=5$
Mode of A = 3; Median of A = 3
Mode of B = 5; Median of B = 7
Mode of C = 4; Median of C = 4
(Mean of A = 3)
$\neq$ (Mode of C = 4)(Mean of C = 5)
$\neq$ (Median of B = 4)(Median of B = 7)
$\neq$ (Mode of A = 3)Mean of A = 3, Mode of A = 3, Median of A = 3