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Measure of Central Tendency (Mean, Median, Quartiles and Mode) — Mathematics STD 10 — Question

ICSE BoardEnglish MediumSTD 10MathematicsMeasure of Central Tendency (Mean, Median, Quartiles and Mode)4 Marks
Question
Using step$- $deviation method , calculate the mean marks of the following distribution.
$C.I$ $50-55$ $55-60 $ $60-65$ $65-70$ $70-75$ $75-80$ $80-85$ $85-90$
Frequency $5$ $20$ $10$ $10$ $9$ $6$ $12$ $8$
✓

Answer

Let the assumed mean $A=72.5$
$C.I$ $f_i$ Mid$-$value
$(x_i)$		 $A=72.5$
$t=(x-A)/(i)$		 $f_it_i$
$50-55$ $5$ $52.5$ $-4$ $-20$
$55-60$ $20$ $57.5$ $-3$ $-60$
$60-65$ $10$ $62.5$ $-2$ $-20$
$65-70$ $10$ $67.5$ $-1$ $-10$
$70-75$ $9$ $72.5$ $0$ $0$
$75-80$ $6$ $77.5$ $1$ $6$
$80-85$ $12$ $82.5$ $2$ $24$
$85-90$ $8$ $87.5$ $3$ $24$
Total $80$     $-56$
$\text { Mean }= A +\frac{ f _{ i } t _{ i }}{ n } \times i$
$=72.5+\left(-\frac{56}{80}\right) \times 5$
$=72.5-0.7 \times 5$
$=72.5-3.5$
$=69$

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