5 Mark Question

Question 15 Marks

Find the missing frequency (p) for the following distribution whose mean is 7.68

x:	3	5	7	9	11	13
f:	6	8	15	p	8	4

Answer

Calculation of mean:

$x_i$	$f_i$	$x_if_i$
3	6	18
5	8	40
7	15	105
9	p	9p
11	8	88
13	4	52
Total	$\sum\text{f}_\text{i}=41+\text{p}$	$\sum\text{f}_\text{i}\text{x}_\text{i}=303+9\text{p}$

We have,
$\sum\text{f}_\text{i}=41+\text{p},\sum\text{f}_\text{i}\text{x}_\text{i}=303+9\text{p}$
$\therefore\text{Mean}=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}$
$\Rightarrow7.68=\frac{303+9\text{p}}{41+\text{p}}$
$\Rightarrow303 + 9\text{p} = 314.88 + 7.68\text{p}$
$\Rightarrow1.32\text{p} = 314.88 - 303$
$\Rightarrow\text{p}=\frac{11.88}{1.32}$
$\Rightarrow\text{p} = 9$

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Question 25 Marks

Find the value of p for the following distribution whose mean is 16.6

x:	8	12	15	p	20	25	30
f:	12	16	20	24	16	8	4

Answer

Calculation of mean:

$x_i$	$f_i$	$x_if_i$
8	12	96
12	16	192
15	20	300
p	24	24p
20	16	320
25	8	200
30	4	120
Total	$\sum\text{f}_\text{i}=\text{N}=100$	$\sum\text{f}_\text{i}\text{x}_\text{i}=1288+24\text{p}$

We have,
$\sum\text{f}_\text{i}=100+\text{p},\sum\text{f}_\text{i}\text{x}_\text{i}=1228+24\text{p}$
$\therefore\text{Mean}=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}$
$\Rightarrow16.6=\frac{1228+24\text{p}}{100}$
$\Rightarrow1228 + 24\text{p} = 16.6 × 100$
$\Rightarrow24\text{p} = 1660 - 1228$
$\Rightarrow\text{p} = 18$

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Question 35 Marks

Find the missing value of p for the following distribution whose mean is 12.58

x:	5	8	10	12	p	20	25
f:	2	5	8	22	7	4	2

Answer

Calculation of mean:

$x_i$	$f_i$	$x_if_i$
5	2	10
8	5	40
10	8	80
12	22	264
p	7	7p
20	4	80
25	2	50
Total	$\sum\text{f}_\text{i}=\text{N}=50$	$\sum\text{f}_\text{i}\text{x}_\text{i}=524+7\text{p}$

We have,
$\sum\text{f}_\text{i}=50,\sum\text{f}_\text{i}\text{x}_\text{i}=524+7\text{p}$
$\therefore\text{Mean}=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}$
$\Rightarrow12.58=\frac{524+7\text{p}}{50}$
$\Rightarrow524 + 7\text{p} = 12.58 × 50$
$\Rightarrow7\text{p} = 629 - 524$
$\Rightarrow\text{p}=\frac{105}{7}$
$\Rightarrow\text{p} = 15$

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Question 45 Marks

Find the value of p, if the mean of the following distribution is 20

x:	15	17	19	20 + p	23
f:	2	3	4	5 p	6

Answer

Calculation of mean:

$x _{ i }$	$f _{ i }$	$\left(f_i x _{ i }\right)$
15	2	30
17	3	51
19	4	76
20 + p	5p	(20 + p) 5p
23	6	138
Total	$\sum\text{f}_\text{i}=15+5\text{p}$	$\sum\text{f}_\text{i}\text{x}_\text{i}=295+(20+\text{p})\ 5\text{p}$

We have,
$\sum\text{f}_\text{i}=15+5\text{p},\sum\text{f}_\text{i}\text{x}_\text{i}=295+(20+\text{p})\ 5\text{p}$
$\therefore\text{Mean}=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}$
$\Rightarrow20=\frac{(295+(20+\text{p})5\text{p})}{15+5\text{p}}$
$\Rightarrow 295+100 p+5 p^2=300+100 p$
$\Rightarrow 5 p^2=300-295$
$\Rightarrow 5 p^2=5$
$\Rightarrow p^2=1$
$\Rightarrow p=1$

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Question 55 Marks

The following table shows the weights of 12 persons.

Weight (in kg):	48	50	52	54	58
Number of persons:	4	3	2	2	1

Find the median and mean weights. Using empirical relation, calculate its mode.

Answer

Weight ( $x _{ i }$ )	48	50	52	54	58	Total
Number of Persons ( $f _{ i }$ )	4	3	2	2	1	$\sum\text{f}_\text{i}=12$
$\left(f_i x _{ i }\right)$	192	150	104	108	58	$\sum\text{f}_\text{i}\text{x}_\text{i}=612$

$\text{mean} = \frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}\frac{612}{12}=51\text{kg.}$
Here, n = 12
Median = value of $\frac{\text{n}}{2}\text{th}$ observation $\frac{\text{n}}{2}+1^\text{th}$ observation
$\Rightarrow\text{Median}=\frac{\text{Value of 6}^\text{th}\ \text{observation}+\text{Value of 7}^\text{th}\ \text{observation}}{2}$
$\Rightarrow\text{Median}=\frac{50+50}{2}=50\text{kg.}$
Now,
Mode = 3 Median - 2 Mean
⇒ Mode = 3 × 50 - 2 × 51
⇒ Mode = 150 - 102
⇒ Mode = 48kg.
Thus, Mean = 51kg, Median = 50 kg and Mode = 48kg.

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