Maths 5 Marks Questions

Now, n = 30
So, $\frac{n}{2} = \frac{{30}}{2} = 15$
This observation lies in the class 55-60,
So, 55-60 is the median class.
Therefore,
l = 55
h = 5
f = 6
cf = 13
$\therefore $ Median $ = l + \left( {\frac{{\frac{n}{2} - cf}}{f}} \right) \times h = 55 + \left( {\frac{{15 - 13}}{{6}}} \right) \times 5$
$ = 55 + \frac{{10}}{6} = 55 + \frac{5}{3}$
= 55 + 1.67 = 56.67
Hence, the median weight of the students is 56.67 kg.

1. N = 100
   Mean = $\frac { \Sigma f_i x_i } { N } = \frac { 832 } { 100 }$ = 8.32
2. $ \frac{N}{2} = \frac{{100}}{2}$ = 50
   The cumulative frequency just greater than $\frac {N} {2}$ is 76, then the median class is 7 - 10 such that
   l = 7, h = 10 - 7 = 3, f = 40, F = 36
   Median = l + $\frac { \frac { N } { 2 } - F } { f } \times$ h
   = 7 + $\frac { 50 - 36 } { 40 } \times$ 3
   $= 7 + \frac {42} {40}$ = 7 + 1.05 = 8.05
3. Mode = 3 Median - 2 Mean
   = 3 $\times$ 8.05 - 2 $\times$ 8.32 = 7.51

Now, n = 40
So, $\frac{n}{2} = \frac{{40}}{2}$ = 20
This observation lies in the class 144.5 - 153.5.
So, 144.5 - 153.5 is the median class.
Therefore,
l = 144.5
h = 9
cf = 17
f = 12
$\therefore $ Median = l + $\left( {\frac{{\frac{n}{2} - cf}}{f}} \right) \times$ h = 144.5 + $\left( {\frac{{20 - 17}}{{12}}} \right) \times$ 9
= 144.5 + 2.25 = 146.75 mm
Hence, the median length of the leaves is 146.75 mm

Now, n = 100
So, $\frac{n}{2} = \frac{{100}}{2}$ = 50
This observation lies in class 35 - 40.
So, 35 - 40 is the median class.
Therefore,
l = 35
h = 5
cf = 45
f = 33
$\therefore $ Median = l + $\left( {\frac{{\frac{n}{2} - cf}}{f}} \right) \times$ h = 35 + $\left( {\frac{{50 - 45}}{{33}}} \right) \times$ 5
= 35 + $ \frac{{25}}{{33}}$ = 35 + 0.76 = 35.76 years
Hence, the median age is 35.76 years.

Here, $\sum f _ { i } = n = 60$, then $\frac { n } { 2 } = \frac { 60 } { 2 } = 30$, also, median of the distribution is 28.5, which lies in interval 20 – 30.
$\therefore$ Median class = 20 – 30
So, l = 20, n = 60, f = 20, cf = 5 + x and h = 10
$\because 45 + x + y = 60$
$\Rightarrow x + y = 15$ ………...........(i)
Now, Median = $l + \left[ \frac { \frac { n } { 2 } - c f } { f } \right] \times h$
$\Rightarrow { 28.5 = 20 + \left[ \frac { 30 - ( 5 + x ) } { 20 } \right] \times 10 }$
$\Rightarrow 28.5 = 20 + \frac { 30 - 5 - x } { 2 }$
$\Rightarrow { 28.5 } = \frac { 40 + 25 - x } { 2 }$
$\Rightarrow 57.0 = 65 - x$
$\Rightarrow x = 65 - 57 = 8$
$\Rightarrow$ x = 8
Putting the value of x in eq. (i), we get,
8 + y = 15
$\Rightarrow$ y = 7
Hence the value of x and y are 8 and 7 respectively.

1. Let a = 135.
   Now, h = 20
   Using the step-deviation method,
   $Mean, \overline x = a + \left( {\frac{{\sum {{f_i}{u_i}} }}{{\sum {{f_i}} }}} \right) \times h$ $ = 135 + \left( {\frac{7}{{68}}} \right) \times 20$
   $ = 135 + \frac{{35}}{{17}}$ = 135 + 2.05 = 137.05
2. Now, N = 68
   So, $\frac{N}{2} = \frac{{68}}{2} = 34$
   This observation lies in class 125-145.
   Therefore, 125-145 is the median class.
   So, l = 125, CF = 22, f = 20
   $\therefore Median = l + \left( {\frac{{\frac{N}{2} - CF}}{f}} \right) \times h$
   $ = 125 + \left( {\frac{{34 - 22}}{{20}}} \right) \times 20$ = 125 + 12 = 137
3. Mode = 3 Median - 2 Mean
   = 3$\times$137 - 2$\times$137.05 = 136.9

We have,
N = $\Sigma f _i$ = 100
$\Rightarrow $ 76 + x + y = 100
$\Rightarrow $ x + y = 24
It is given that the median is 525. Clearly, it lies in the class 500 - 600
$\therefore$ l = 500, h = 100, f = 20, F = 36 + x and N = 100
Now, Median = l + $\frac { \frac { N } { 2 } - F } { f } \times$ h
$\Rightarrow$ 525 = 500 + $\frac { 50 - ( 36 + x ) } { 20 } \times$ 100
$\Rightarrow $ 525 - 500 = (14 - x)5
$\Rightarrow $ 25 = 70 - 5x
$\Rightarrow $ 5x = 45
$\Rightarrow $ x = 9
Putting x = 9 in x + y = 24, we get y = 15
Hence, x = 9 and y = 15