Tag: assumed mean method of finding mean

Questions Related to assumed mean method of finding mean

Sum of squares of deviation of $10$ observations measured from $5$ is $17$ and sum of squares of observations is $170$ then mean of observation is

maths measure of central tendency assumed mean method assumed mean method of finding mean mean and median

  1. $40.3$

  2. $4.5$

  3. $4$

  4. $4.03$

  5. $4.3$

Show answer
Correct Option: D
Explanation:

$x _1,x _2,x _3,x _4...x _10\x _1^2+x _2^2+x _3^2+x _4^2+....x _10^2=17 (Given) \rightarrow (i)$

$(x _1-5)^2+(x _2-5)^2+(x _3-5)^2+....+(x _10-5)^2=17(Given)\rightarrow (i)$
$Mean (M)=\cfrac{x _1+x-2+x _3+....x _10}{10}\ \Rightarrow x _1+x _2+x _3+....+x _10=10M\rightarrow(iii)$
From equation $(ii)$ we get
$x _1^2+25-10x _1+x _2^2+25-10x _2+x _3^2+25-10x _3+...+x _10^2+25-10x _10=17\ \Rightarrow (x _1^2+x _2^2+....x _10^2)-10(x _1+x _2+x _3+....+x _10)+25\times10=17\ \Rightarrow 170-10(10M)+250=17\ \Rightarrow420-100M=17\ \Rightarrow100M=403\ \Rightarrow M=\cfrac{403}{100}=4.03$

The sum of the deviations of a set of values $x 1, x _2$, ...... $x _n$ measured from $50$ is $-10$ and the sum of deviations of the values from $46$ is $70$. The mean is __________.

maths measure of central tendency assumed mean method assumed mean method of finding mean mean and median

  1. $49$

  2. $49.5$

  3. $49.75$

  4. $50$

Show answer
Correct Option: B
Explanation:

Sum of deviations from $50$ is $-10$


$\Rightarrow \sum(xi - 50) = -10$
     $\sum x _i - 50\sum1 = -10$
     $\sum x _i - 50n = -10$

$\therefore y-50n=-10.....(1)$

Sum of deviations from $46$ is $70$

$\Rightarrow \sum(x _i - 46) = 70$
     $\sum x _i - 46\sum1 = 70$
     $\sum x _i - 46n = 70$

$\therefore y-46n=70.....(2)$


Solving $(1)$ and $(2)$, we get
$4n = 80$ i.e. $n=20$

Putting value of $n$ in $(1)$, we get
$y=990$

Mean $= \dfrac{\sum x _i}{n} = \dfrac{y}{n} = \dfrac{990}{20} = 49.5$

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?

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

  1. $109$

  2. $106$

  3. $150$

  4. $160$

Show answer
Correct Option: B
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

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

  1. $51.12$ litres

  2. $54.12$ litres

  3. $57.12$ litres

  4. $60.12$ litres

Show answer
Correct Option: C
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.