Michael Beale Google Chrome Secure https/www.mathud.com/Student/PlayerHomework.a

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Michael Beale Google Chrome Secure https/www.mathud.com/Student/PlayerHomework.aspxthomeworkdd-4646949088questionld 38flushed-falseicld 4838257centerwin y Stat 2 Spring 2018 Michael Bae 2/1/18 12:52 PM Homework: HW 1 Normal Distrib, Sampling Distrib Ch 6.2 & Ch Score: 04 of 1 pt Score: 79.59%, 5.57 of 7 7.3.15-T Question Hep A survey roates that 20%of cor surrers tom a ce tan regon wilngto spend etfor products d servces hom socialy responsatie com panes he nesegders dere these consumers as socially conscious consumers Acconding to one of the setearchers, markrters need to know who these consumers ase if they want to maximire the social and business retuarm of ther cause marketing efforts Suppose a sampke of 100 consumers fhom the segion is selected If possible, use the normal ds-bution to approximate the sampling distaibution of the proportion Compiete parts (a) trough ig)edow The potatity 6 % Round to thewo deciemal places as needed ) b.wtut is the prtataity that nte sagie between 21% and 31%are wilngtospendetratr products and servceston soul, resporsie com-' the prestany is 7458% (Round to two decimal places as needed s. What is the probabsity that in the sample more than 2P% ae wilieng to spend extra for products and services trom soculty responsble comparies? The protatinys ara Flound to two decienal places as needed) Two probable, es 50% Round to two decoimal places as needed) The prebabity io Round 10 two deoima places as reeded) u More Enter your answer in he answer box and then cick Check A

Explanation / Answer

e)std error of mean =(p(1-p)/n)1/2 =(0.26*(!-0.26)/400)1/2 =0.0219

hence probability =P(0.21<P<0.31)=P((0.21-0.26)/0.0219<Z<(0.31-0.26)/0.0219)=P(-2.2798<Z<2.2798)

=0.9887-.0113 =0.9774 ~ 97.74%

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