Solved 1 Answer
See More Answers for FREE
Enhance your learning with StudyX
Receive support from our dedicated community users and experts
See up to 20 answers per week for free
Experience reliable customer service
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 ...