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.
$x̄$ =$a$+$h({Σf_i u_i }/{Σf_i})$
=$57$+$2({37.5 }/{400})$
=$57$+$0.1875$
=$57.1875$