Exercises 113 Data: C1: -3 -2 -1 0 1 2 3 abmve E.F. Dialog box: Session command: Calc > Probability Distributions > Normal MTB CDF C1; SUBC> Normal 0 1. Choose Cumulative probability. Choose Input...

Can you help with 4.7.5 please classes are cancelled due to virus, there is no way I can ask teacher or students at school for help we are not allowed on campus

Extracted text: Exercises 113 Data: C1: -3 -2 -1 0 1 2 3 abmve E.F. Dialog box: Session command: Calc > Probability Distributions > Normal MTB CDF C1; SUBC> Normal 0 1. Choose Cumulative probability. Choose Input column and type Cl. Click OK. Output: Cumulative Distribution Function Normal with mean =0 and standard deviation = 1.00000 %3D х P( X <= x)="" -3.0000="" 0.0013="" -2.0000="" 0.0228="" -1.0000="" 0.1587="" 0.0000="" 0.5000="" 1.0000="" 0.8413="" 2.0000="" 0.9772="" 3.0000="" 0.9987="" figure="" 4.7.4="" minitab="" calculation="" of="" cumulative="" standard="" normal="" probabilities.="" (b)="" exercises="" 4.7.1="" for="" another="" subject="" (a="" 29-year-old="" male)="" in="" the="" study="" by="" diskin="" et="" al.="" (a-11),="" acetone="" sal="" levels="" were="" normally="" distributed="" with="" a="" mean="" of="" 870="" and="" a="" standard="" deviation="" of="" 211="" ppb.="" find="" the="" probability="" that="" on="" a="" given="" day="" the="" subject's="" acetone="" level="" is:="" (a)="" between="" 600="" and="" 1000="" ppb="" (b)="" over="" 900="" ppb="" (c)="" under="" 500="" ppb="" (d)="" between="" 900="" and="" 1100="" ppb="" 4.7.2="" in="" the="" study="" of="" fingerprints,="" an="" important="" quantitative="" characteristic="" is="" the="" total="" ridge="" count="" for="" the="" 10="" fingers="" of="" an="" individual,="" suppose="" that="" the="" total="" ridge="" counts="" of="" individuals="" in="" a="" certain="" population="" are="" approximately="" normally="" distributed="" with="" a="" mean="" of="" 140="" and="" a="" standard="" deviation="" of="" 50,="" find="" the="" probability="" that="" an="" individual="" picked="" at="" random="" from="" this="" population="" will="" have="" a="" ridge="" count="" of:="" (a)="" 200="" or="" more="" (b)="" less="" than="" 100="" (c)="" between="" 100="" and="">


Extracted text: (d) Between 200 and 250 (e) In a population of 10,000 people how many would you expect to have a ridge connt of 200 or more? 4.7.3 One of the variables collected in the North Carolina Birth Registry data (A-3) is pounde gained during pregnancy. According to data from the entire registry for 2001, the number of pounds gained during pregnancy was approximately normally distributed with a mean of 30.23 pounds and a standard deviation of 13.84 pounds. Calculate the probability that a randomly selected mother in North Carolina in 2001 gained: (a) Less than 15 pounds during pregnancy (b) More than 40 pounds (c) Between 14 and 40 pounds (d) Less than 10 pounds (e) Between 10 and 20 pounds ughuo ielum 4.7.4 Suppose the average length of stay in a chronic disease hospital of a certain type of patient is 60 days with a standard deviation of 15. If it is reasonable to assume an approximately normal distribution of lengths of stay, find the probability that a randomly selected patient from this group will have a length of stay: (a) Greater than 50 days (b) Less than 30 days (c) Between 30 and 60 days (d) Greater than 90 days 0000. E- 0000.S- 0000 1- 4.7.5 If the total cholesterol values for a certain population are approximately normally dis- tributed with a mean of 200 mg/100 ml and a standard deviation of 20 mg/100 ml, find the probability that an individual picked at random from this population will have a cholesterol value: (a) Between 180 and 200 mg/100 ml (b) Greater than 225 mg/100 ml (c) Less than 150 mg/100 ml (d) Between 190 and 210 mg/100 ml