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Question Number 101062 by bemath last updated on 30/Jun/20

Answered by MJS last updated on 30/Jun/20

$$\mathrm{after}‘‘\mathrm{many}{\mathrm{times}}^{″}\mathrm{implies}\mathrm{that}$$$$p=\frac{10}{25}=\frac{2}{5}\pm 0$$$$\Rightarrow \mathrm{there}\mathrm{are}\frac{25}{p}=62.5\approx 60\mathrm{candies}\mathrm{in}\mathrm{the}\mathrm{jar}$$

Commented by bemath last updated on 30/Jun/20

$$\mathrm{sir}\mathrm{why}\frac{2}{5}\pm 0?$$

Commented by bemath last updated on 30/Jun/20

$$\mathrm{how}\mathrm{sir}\mathrm{get}62.5?$$

Commented by bemath last updated on 30/Jun/20

$$\frac{25}{\left(\frac{2}{5}\right)}=\frac{125}{2}=62.5\approx 60$$

Commented by MJS last updated on 30/Jun/20

$$\mathrm{if}\mathrm{we}\mathrm{do}\mathrm{not}\mathrm{know}\mathrm{the}\mathrm{probability}\mathrm{and}\mathrm{we}$$$$\mathrm{make}\mathrm{one}\mathrm{test}\mathrm{as}\mathrm{described}\mathrm{here},\mathrm{we}\mathrm{get}\mathrm{a}$$$$\mathrm{normally}\mathrm{distributed}p.\mathrm{we}\mathrm{then}\mathrm{have}\mathrm{to}$$$$\mathrm{define}\mathrm{a}\mathrm{confidence}\mathrm{interval}.\mathrm{usually}\mathrm{you}$$$$\mathrm{want}\mathrm{to}\mathrm{be}\mathrm{at}\mathrm{least}95\mathrm{\%}\mathrm{sure}\mathrm{that}\mathrm{the}$$$$\mathrm{probability}\mathrm{lies}\mathrm{within}[p-ϵ;p+ϵ]$$$${\mathrm{you}}^{\prime }\mathrm{ll}\mathrm{have}\mathrm{to}\mathrm{study}\mathrm{this}.$$$$\mathrm{but}\mathrm{after}‘‘\mathrm{many}{\mathrm{tests}}^{″}\mathrm{means}\mathrm{we}\mathrm{are}100\mathrm{\%}$$$$\mathrm{sure}\mathrm{the}\mathrm{probability}\mathrm{is}\frac{2}{5}.\mathrm{then}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{easy}:$$$$\mathrm{we}\mathrm{marked}25$$$$\mathrm{if}\mathrm{we}\mathrm{take}\mathrm{out}25\mathrm{we}\mathrm{get}10\mathrm{marked}$$$$\mathrm{we}\mathrm{must}\mathrm{take}\mathrm{out}2.5\times 25\mathrm{to}\mathrm{get}2.5\times 10\mathrm{marked}$$$$2.5\times 25=62.5$$

Commented by bemath last updated on 30/Jun/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}\mathrm{for}\mathrm{your}\mathrm{explanation}$$

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