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Statistics
NCERT Exemplar Ch 12
CBSE_NCERT_EXEMPLAR_CH12
Grade 10

Question:

Find the mean of the following distribution using the Assumed Mean Method:
Class: 10-25, 25-40, 40-55, 55-70, 70-85, 85-100
Frequency: 2, 3, 7, 6, 6, 6

Step-by-Step Solution

Key Concept: Class marks $x_i$: 17.5, 32.5, 47.5, 62.5, 77.5, 92.5. Let $a = 62.5$. $d_i = x_i - a$. $\bar{x} = a + \dfrac{\sum f_i d_i}{\sum f_i}$.
Stepwise Solution:

Class marks $x_i$: 17.5, 32.5, 47.5, 62.5, 77.5, 92.5. Assumed mean $a = 62.5$. [0.5 Mark]

$d_i = x_i - 62.5$: $-45, -30, -15, 0, 15, 30$. [1.0 Mark]

$f_i d_i$: $-90, -90, -105, 0, 90, 180$. $\sum f_i d_i = -15$. $\sum f_i = 30$. [1.0 Mark]

$\bar{x} = 62.5 + \dfrac{-15}{30} = 62.5 - 0.5 = 62$. Mean is $62$. [0.5 Mark]

Marking Scheme:

• Class marks and deviations $d_i$: 1.5 Marks
• Sum $\sum f_i d_i = -15$ and $\sum f_i = 30$: 1.0 Mark
• Calculating mean $= 62$: 0.5 Mark

Correct Answer:
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