Mean for the frequency distribution is 1.46 then choose the correct options -
No. of Accidents
0
1
2
3
4
5
Total
Frequency (f)
46
$f_{1}$
$f_{2}$
25
10
5
200

f_{1} = 76

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f_{2} = 45

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f_{1} = 70

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f_{2} = 38

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Solution

The correct options are
A $f_{1} = 76$
D $f_{2} = 38$
The calculation are as under :
$x_{i}$
$f_{i}$
$f_{i} x_{i}$
0
46
0
1
$f_{1}$
$f_{1}$
2
$f_{2}$
$2 f_{2}$
3
25
75
4
10
40
5
5
25

$N = \sum f_{1} = 86 + f_{1} + f_{2}; \sum f_{i} x_{i} = 140 + f_{1} 2 f_{2}$
It isgiven that N $=$ 200.
Therefore, $86 + f_{1} + f_{2} = 200 \Rightarrow f_{1} + f_{2} = 114$ ..... (1)
Now, $¯ x = 1.46 \Rightarrow 1.46 = \frac{\sum f_{i} x_{i}}{N}$
$\Rightarrow 1.46 = \frac{140 + f_{1} + 2 f_{2}}{200}$
$\Rightarrow 292 = 140 + f_{1} + 2 f_{2} \Rightarrow f_{1} + 2 f_{2} = 152$ ...... (2)
Solving (1) and (2), we get $f_{1} = 76$ and $f_{2} = 38$

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∣ ∣ ∣_{1} \times_{2} ∣ ∣ ∣ =_{1} ._{2}

∣ ∣ ∣_{1} +_{2} ∣ ∣ ∣

f_{1}

f_{2}

f_{3}

f_{1}

f_{2}

Class	0 - 20	20 - 40	40 - 60	60 - 80	80 - 100	100 - 120	Total
Frequency	5	$f_{1}$	10	$f_{2}$	7	8	50

50

f_{1}

f_{2}

Class	0-20	20-40	40-60	60-80	80-100	Total
Frequency	17	28	$f_{1}$	24	$f_{2}$	120

No. of accidents	0	1	2	3	4	5
No. of days	46	$f_{1}$	$f_{2}$	25	10	5

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No. of Accidents	0	1	2	3	4	5	Total
Frequency (f)	46	$f_{1}$	$f_{2}$	25	10	5	200