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.



1
Expert's answer
2022-04-06T16:06:15-0400

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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