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Question Number 33822 by mondodotto@gmail.com last updated on 25/Apr/18

The following table shows the  distributuons of 100 families according  to their expenditure per week.  The mode is given to be 24.  ∣((expenditure)/(Number of families))∣((10−20)/x)∣((20−30)/(27))∣((30−40)/y)∣((40−50)/(15))∣  (a) calculate the missing frequency  (b)calculate the mean  (c)calculate the median

$$\boldsymbol{\mathrm{The}}\:\boldsymbol{\mathrm{following}}\:\boldsymbol{\mathrm{table}}\:\boldsymbol{\mathrm{shows}}\:\boldsymbol{\mathrm{the}} \\ $$$$\boldsymbol{\mathrm{distributuons}}\:\boldsymbol{\mathrm{of}}\:\mathrm{100}\:\boldsymbol{\mathrm{families}}\:\boldsymbol{\mathrm{according}} \\ $$$$\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{their}}\:\boldsymbol{\mathrm{expenditure}}\:\boldsymbol{\mathrm{per}}\:\boldsymbol{\mathrm{week}}. \\ $$$$\boldsymbol{\mathrm{The}}\:\boldsymbol{\mathrm{mode}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{given}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{be}}\:\mathrm{24}. \\ $$$$\mid\frac{\boldsymbol{\mathrm{expenditure}}}{\boldsymbol{\mathrm{Number}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{families}}}\mid\frac{\mathrm{10}−\mathrm{20}}{\mathrm{x}}\mid\frac{\mathrm{20}−\mathrm{30}}{\mathrm{27}}\mid\frac{\mathrm{30}−\mathrm{40}}{\boldsymbol{\mathrm{y}}}\mid\frac{\mathrm{40}−\mathrm{50}}{\mathrm{15}}\mid \\ $$$$\left(\boldsymbol{\mathrm{a}}\right)\:\boldsymbol{\mathrm{calculate}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{missing}}\:\boldsymbol{\mathrm{frequency}} \\ $$$$\left(\boldsymbol{\mathrm{b}}\right)\boldsymbol{\mathrm{calculate}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{mean}} \\ $$$$\left(\boldsymbol{\mathrm{c}}\right)\boldsymbol{\mathrm{calculate}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{median}} \\ $$

Commented by math1967 last updated on 25/Apr/18

There is some error.Here mode is 24  ∴modal class 20−30∴x+y<2×27  but here x+y=100−42=58>2×27

$${There}\:{is}\:{some}\:{error}.{Here}\:{mode}\:{is}\:\mathrm{24} \\ $$$$\therefore{modal}\:{class}\:\mathrm{20}−\mathrm{30}\therefore{x}+{y}<\mathrm{2}×\mathrm{27} \\ $$$${but}\:{here}\:{x}+{y}=\mathrm{100}−\mathrm{42}=\mathrm{58}>\mathrm{2}×\mathrm{27} \\ $$

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