
#1
Mar512, 09:09 PM

P: 2

Hi Physics Forum! I hope this is the right section... I couldn't find a section on statistics.
This is a rather easy question but I can't seem to get the answer that the answer key says! Average Life expectancy μ=72 standard Deviation ( in years ) δ=5 The question: Recent studies have suggested that advances in healthcare may increase the lifespan of employees. How much would the average life expectancy need to rise in order for 70% of people to live to age 70. http://miha.ef.unilj.si/_dokumenti3...ormtables.pdf (Normal Distribution table in case you guys dont have !) I don't know exactly how to approach this question. When it says X percent live up to Y age. Does that mean that the spread from the mean is X percent? Or up until Y age it is X percent. If we try the former then that means 70 pecent of the empolyees live between the ages 66.85 and 77.15 ...but that doesn't seem like the right way to go because the answer has only one number. If we try the latter that means 70 percent of the employees live from 074.6 (0.52 Z scores away from the mean) But 74.6 is already higher than 70 which is what we are trying to get at... I'm so lost can anybody help out? Thanks 



#2
Mar512, 09:41 PM

HW Helper
Thanks
P: 4,664

RGV 



#3
Mar512, 09:59 PM

P: 2

sir what is this equation? Could u please explain it for me? 



#4
Mar512, 10:50 PM

P: 40

Normal Probability Distributions! :earlier when you were got the 74. something years, that meant that 70% of people live 0 to 74.whatever years. 


Register to reply 
Related Discussions  
Probability Theory Standard Normal Distributions  Precalculus Mathematics Homework  3  
Probability Question  Nonstandard Normal Distributions  Calculus & Beyond Homework  3  
Normal Distributions  General Math  4  
Probability : joint density function of 3 Normal Distributions  Precalculus Mathematics Homework  3  
Normal and power law distributions  General Math  1 