Question #324417

6. According to a study from 2012, it was found that 19% of people use their browser in private mode. For a new study on privacy policies, you will consider a random sample of 200 people. Let X be the number of people who use their browser in private mode.

(a) What is the expected value (mean) and standard deviation for X?

(b) Use a normal approximation to estimate the probability that you will find 50 or more people using their browser in private mode in your sample.

(c) Use a normal approximation to estimate the probability that you will find more than 35 and up to 45 people (inclusively) using their browser in private mode in your sample.



Expert's answer

a)  μ=np=200(0.19)=38,σ=np(1p)=200(0.19)(0.81)5.55Using  normal  approximation  b)  P(x50)=P(x49.5)=P(z49.5385.55)=P(z2.07)=0.5P(0<z<2.07)=0.50.4808=0.0192c)  P(35<x<45)=P(34.5<x<45.5)=P(34.5385.55<z<45.5385.55)=P(0.63<z<1.35)=P(0<z<0.63)P(0<z<1.35)=0.2357+0.4115=0.6472a)\;\mu =n p=200(0.19)=38,\\ \sigma= \sqrt{np(1-p)}= \sqrt{200(0.19)(0.81)}\approx5.55\\ Using\; normal\; approximation\;\\ b)\;P(x\geq 50)=P(x\geq 49.5)\\ =P(z\geq \frac{49.5-38}{5.55})\\ =P(z\geq 2.07)=0.5-P(0<z<2.07)\\ =0.5-0.4808=0.0192\\ c)\;P(35<x< 45)=P(34.5<x<45.5)\\ =P(\frac{34.5-38}{5.55}<z<\frac{45.5-38}{5.55})\\ =P(-0.63<z<1.35)\\ =P(0<z<0.63)-P(0<z<1.35)\\ =0.2357+0.4115=0.6472\\


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