watch online Game of T xV8Gorillatid Just watch it! x y VA HW eg (Sections 5.3-5

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watch online Game of T xV8Gorillatid Just watch it! x y VA HW eg (Sections 5.3-5.5) x C Chegg study l Guided se x C www.webassign net/web/Student/Assignment Responses/Ma 15985793 Apps Bookmarks GUITAR TABS AND CH Make a payment B Courses Blackboard Mail Rashid, Umer D matlab -a-practica-in Membe Area wepc D virginia Community C M Inbox (3) 241 Applicati IJob op Need Help? Read It Talk to Tutor -1 points DevoreStat9 5.E.051. 0130 Submissions Used My Notes. Ask Your Teacher Time taken by a randomly selected applicant for a mortgage to fill out a certain form has a normal distribution with mean value 10 min and standard deviation 2 min. If five individuals fill out a form on one day and six on another, what is the probability that the sample average amount of time taken on each day is at most 11 min? (Round your answer to four decimal places.) You may need to use the appropriate table in the Appendix of Tables to answer this question. Need Help? Read It Talk to a Tutor 5. 2 points Devo restat9 5.E.053. 030 Submissions Used My Notes Ask Your Teacher Rockwell hardness of pins of a certain type is known to have a mean value of 50 and a standard deviation of 1.7. Round your answers to four decimal places.) (a) If the distribution is normal, what is the probability that the sample mean hardness for a random sample of 10 pins is at least 51? (b) What is the (approximate) probability that the sample mean hardness for a random sample of 43 pins is at least 51? You may need to use the appropriate table in the Appendix of Tables to answer this question. :05 AM

Explanation / Answer

4) for n=5; std error =std deviation/(n)1/2=0.8944

hence P(X<11)=P(Z<(11-10)/0.8944)=P(Z<1.118)=0.8682

for n=6; std error =std deviation/(n)1/2=0.8165

hence P(X<11)=P(Z<(11-10)/0.8165)=P(Z<1.2247)=0.8897

hence probability that on each day average is at most 11 =0.8682*0.8897=0.7724

5) for n=10; std error =std deviation/(n)1/2=0.5376

P(X>51)=1-P(Z<(51-50)/0.5376)=1-P(Z<1.8602)=1-0.9686=0.0314

for n=43; std error =std deviation/(n)1/2=0.2592

P(X>51)=1-P(Z<(51-50)/0.2592)=1-P(Z<3.8573)=1-0.99994=0.00006

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