5. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained
varying number of mangoes.The following was the distribution of mangoes according to the number of boxes.
Number of mangoes | Number of boxes |
50-52 | 15 |
53-55 | 110 |
56-58 | 135 |
59-61 | 115 |
62-64 | 25 |
Solution:
Therefore frequency table is given by:
Number of mangoes | Number of boxes ($f_i$) | Midpoint($x_i$) | $d_i=x_i-57$ | $u_i= {x_i-a}/{h}$ | $f_ix_i$ | $f_id_i$ | $f_iu_i$ |
50-52 | 15 | 51 | -6 | -3 | 765 | -90 | -45 |
53-55 | 110 | 54 | -3 | -1.5 | 5940 | -330 | -165 |
56-58 | 135 | 57 | 0 | 0 | 7695 | 0 | 0 |
59-61 | 115 | 60 | 3 | 1.5 | 6900 | 345 | 172.5 |
62-64 | 25 | 63 | 6 | 3 | 1575 | 150 | 75 |
Total | 400 | | | | 22875 | 75 | 37.5 |
Let assumed mean of $x_i$, $a$ = 57 and class size $h$ = 2
The Direct Method: $x̄$ = ${∑f_ix_i}/{∑f_i}$
= $22875/400$
=$57.1875$
The Assumed Mean Method: $x̄$ = $a$ +${Σf_i d_i }/{Σf_i}$
= $57$ +${75 }/{400}$
= $57$ +$0.1875$
= $57.1875$
The Step Deviation Method:
$x̄$ =$a$+$h({Σf_i u_i }/{Σf_i})$
Let $u_i= {x_i-a}/{h}$
where $a$ is the assumed mean and $h$ is the class size.
$x̄$ =$a$+$h({Σf_i u_i }/{Σf_i})$
=$57$+$2({37.5 }/{400})$
=$57$+$0.1875$
=$57.1875$