Tag: measures of central tendency

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Questions Related to measures of central tendency

Multiple choice statistics measures of central tendency geometric and harmonic mean geometric mean mean

The mean of the following frequency distribution is 62.8 and the sum of all the frequencies is 50. Compute the missing frequency $\displaystyle f _{1}$ and $\displaystyle f _{2}$.

Class 0-20 20-40 40-60 60-80 80-100 100-120
Frequency 5 $\displaystyle f _{1}$ 10 $\displaystyle f _{2}$ 7 8
  1. $5, 8$
  2. $6, 12$
  3. $8, 11$
  4. $8, 12$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Class       Frequency(f)  ClassMark (x)         fx
0-20         5      10         50
20-40     ${f} _{1} $      30       $ 30{f} _{1} $
40-60         10       50          500
60-80      ${f} _{2} $        70        $ 70{f} _{2} $
80-100          7         90          630
100-120           8         110           880
Total $30 + {f} _{1} +{f} _{2} $   $ 2060  + 30{f} _{1}+70{f} _{2} $

Given $ 30 + {f} _{1} +{f} _{2} = 50 $
$ => {f} _{1} +{f} _{2} = 20 $   -- (1)

Given, Mean $ = \cfrac { \sum { fx }  }{ \sum { f }} =62.8 $
$ => \cfrac { 2060  + 30{f} _{1}+70{f} _{2}}{30 + {f} _{1} +{f} _{2}} = 62.8 $

$ =>  2060  + 30{f} _{1}+70{f} _{2} = 1884 + 62.8{f} _{1} + 62.8{f} _{2} $ 

$ 32.8{f} _{1} - 7.2{f} _{2} =176 $

=> $ 8.2{f} _{1} - 1.8{f} _{2} = 44 $

=> $ 4.1{f} _{1} - 0.9{f} _{2} = 22 $ -- (2)

Solving both equations 1, 2, we get
$ {f} _{1} = 8, {f} _{2} = 12 $

Multiple choice maths measures of central tendency assumed mean method assumed mean method of finding mean mean and median

The exam scores of all 500 students were recorded and it was determined that these scores were normally distributed. If Jane's score is 0.8 standard deviation above the mean, then how many, to the nearest unit, students scored above Jane?

  1. $109$
  2. $106$
  3. $150$
  4. $160$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let m be the mean and s be the standard deviation and find the z score.
$z = (x - m) /s = (0.8 s + m - m) / s = 0.8$
The percentage of student who scored above Jane is (from table of normal distribution).
1 - 0.7881 = 0.2119 = 21.19%
The number of student who scored above Jane is (from table of normal distribution).
21.19% 0f 500 = 106

Multiple choice maths measures of central tendency assumed mean method assumed mean method of finding mean mean and median

Following table gives frequency distribution of milk (in litres) given per week by 50 cows.
Find average (mean) amount of milk given by a cow by 'shift of origin method'.

Milk (in litres) 24 - 30 30 - 36 36 - 42 42 - 48 48 - 54 54 - 60 60 - 66 66 - 72  72 - 78 78 - 84 84 - 90
No. of cows 1 3 8 5 5 5 8 4 6 2 3
  1. $51.12$ litres
  2. $54.12$ litres
  3. $57.12$ litres
  4. $60.12$ litres
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Consider the following table, to calculate mean by "shift of origin method":

$x _i$=mid value of class interval
Assumed mean $a=57$

 $ci$  $f _i$  $x _i$ $d _i=x _i-a$  $f _id _i$
24-30  1 27  27-57= -30  -30
30-36  3 33  33-57= -24  -72
36-42  8 39  39-57= -18  -144
42-48   5 45   45-57= -12  -60
48-54   5 51  51-57= -6  -30
54-60   5 57  57-57=0  0
60-66   8 63   63-57=6   48
66-72   4 69  69-57=12  48
72-78   6 75   75-57=18  108
78-84  2 81   81-57=24  48
84-90 3  87   87-57=30  90
 $N=\Sigma f _i=50$          
 $\Sigma f _id _i=6$

Mean $\overline x=a +\dfrac {\Sigma f _id _i}{N}$
$\therefore \overline x=57 + \dfrac{6}{50}=57.12$

Average amount of milk given by cow is $ 57.12$ litres
Hence, option $C$ is correct.