19. -3 polnts My Notes Ask Your Teacher Reference: Chapter 5.1 Summary Here is a

Question

19. -3 polnts My Notes Ask Your Teacher Reference: Chapter 5.1 Summary Here is a simpile pobability model for multipe-choice tests. Sus tha rach sudent has prahahility p of rrectly answering a question chosen arom universe nf poessile questions. (A stron udent has a igher p than woak student.) The correctncss of answcrs to diffcrent qucstions are indcpondent. l tudent for whom p0.76. Jodi is a good s Use 1 decimal places (a) Use the normal approximation to find the probability that Jodi scores /1% or lower on a 100-question test. (h)Tf the test mrtains 250 questions, what is the prhability that loci will smre 71% or lower? (c) How many questions must the test contain in order to reduce the standard deviation of Jodi's proportion of correct answers to half its value for a 100-itam test? deviatbion of Jodi's proportion of correct 20. + 1 points My Notes Ask Your Teacher A sari It survey interviews nn SRS of 714 rn lege wormen. Suri rise hat 659 of all rn ene wormen hav, been n diet within t e, ast 17 months what is the probability that 77% ar mare a he Women In hesarr le have bnen cri diet? Use 4 decimal places. Submit Answer | Save Pro ress

Explanation / Answer

19) here p=0.76

for n=100 ; std error =(p(1-p)/n)1/2 =0.0427

hence P(X<0.71) =P(Z<(0.71-0.76)/0.0427)=P(Z<-1.1707)=0.1209

b)

for n=250 ; std error =(p(1-p)/n)1/2 =0.0270

hence P(X<0.71) =P(Z<(0.71-0.76)/0.0270)=P(Z<-1.8511)=0.0321

c)as std deviaiton is inversely proportional to sample size ; therefore test must contain 400 questions

20)

here p=0.65

for n=214 ; std error =(p(1-p)/n)1/2 =0.0326

hence probability=P(X>0.77)=1-P(X<0.77)=1-P(Z<(0.77-0.65)/0.0326)=1-P(Z<3.6804)=1-0.9999 =0.0001

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