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10 . Munthly rent for a one becloconn apartment has mean \mu=52200 stendard deviction \tilde{\theta}=\$ 290The selected sample s/ze n=70.Aciording 10 Central limit theorem, 1f population cuidh meeny, and standerad devicetion 6 and take sutficiently lersge reindem - Sample toom the papuleciton, flim the, distobbection of sample mecir) cuill be approximately noramell distributbdwith mean\begin{array}{l}\mu_{\bar{x}}=E[\bar{x}]=11=2200 \\\delta \bar{x}=\sigma / \sqrt{\eta}=\frac{290}{\sqrt{70}}=3 \% .662 . \\\text { Thes } \bar{x} \text { जN }(2200,311.662)\end{array}Thus expected value =2200\text { , } \rho \text { randate } a x 0 s=\frac{\sigma}{\sqrt{n}}=31 \% 662 \text {. }The shape of wes distribution is moomeel(b) The probability of selecing asample of 70 one. bed apperst ment. and finding tece mocnttobe at lecist s2100 per month\begin{array}{l}=P(\bar{x} \geqslant 2100)=P\left(\frac{\bar{x}-11 \bar{x}}{0 \bar{x}} \geqslant \frac{2100-2200}{311.662}\right) \\=P\left(z>\frac{-100}{31.662}\right)=P(z \geqslant-2.8850) \\=P(2 \leqslant 2.585)=0.9981\end{array}Thee \quad P(\bar{x} \geqslant 2100)=0.9181 ...