NCERT Class 10 Mathematics

266 MATHEMATICS

We have, ui = i

x a

h

Therefore, u =

()

1

i i

ii i

ii

xa

f fx a f

h

fhf

✁

✂ ✄✂ ✁ ✂ ☎

✆ ✝ ✞

✂ ✟ ✂ ✠

=

(^1) ii i
ii
fx a f
hf f

✡☞ ☞ ☛

✍ ✌ ✎

✏☞ ☞ ✑

= ✒ ✓

1

x a

h

✔

So, hu = x✕a

i.e., x =a + hu

So, x = ii
i

fu

ah

f

✖✘ ✗

✙ ✚ ✛

✜✘ ✢

Now, substituting the values of a, h, ✣fiui and ✣fi from Table 14.5, we get

x =

29

47.5 15

30

✦ ✧✤ ✥

★✪ ✩✫

= 47.5 + 14.5 = 62

So, the mean marks obtained by a student is 62.

The method discussed above is called the Step-deviation method.

We note that :

✬ the step-deviation method will be convenient to apply if all the di’s have a

common factor.

✬ The mean obtained by all the three methods is the same.

✬ The assumed mean method and step-deviation method are just simplified

forms of the direct method.

✬ The formula x = a + hu still holds if a and h are not as given above, but are

any non-zero numbers such that ui = i

x a

h

✔

.

Let us apply these methods in another example.