Normal Probability Distributions :

[Sixdaysgrace]

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.uni-lj.si/_dokumenti3plus2/195166/norm-tables.pdf
(Normal Distribution table in case you guys don't 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 0-74.6 (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

 

[Ray Vickson]

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.uni-lj.si/_dokumenti3plus2/195166/norm-tables.pdf
(Normal Distribution table in case you guys don't 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 0-74.6 (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

Under the current data, 70 is just 2/5 of a standard deviation below the mean (72), so the probability of a person surviving past 70 is $1 - \Phi(-2/5) \approx 0.655,$ a bit below the required 0.7. So, to what must you increase $\mu$ (the mean) in order to have $P\{ \text{Age} \geq 70 \} = 0.7$? Assume the standard deviation remains the same, since that is what the question implied.

RGV

 

[Sixdaysgrace]

Under the current data, 70 is just 2/5 of a standard deviation below the mean (72), so the probability of a person surviving past 70 is $1 - \Phi(-2/5) \approx 0.655,$ a bit below the required 0.7. So, to what must you increase $\mu$ (the mean) in order to have $P\{ \text{Age} \geq 70 \} = 0.7$? Assume the standard deviation remains the same, since that is what the question implied.

RGV

$1 - \Phi(-2/5) \approx 0.655,$

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

 

[80past2]

1−Φ(−2/5)≈0.655,

Φ is the normal cdf function. Φ(x) gives you the area to the left of x in a normal distribution. since the whole area is 1, and Φ(-2.5) = .345, then the area to the right of -2/5 is .655.

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