Find P(X<= 36.7) using CLT and Z-Scores

[m00nbeam360]

Hi there,

Not sure if this is the right place, but the problem states that a battery has a population mean of 40 hours and standard deviation of 5. Let X represent the mean lifetime of batteries in a simple random sample size of 100. What is P(X <= 36.7)?

I tried computing this with X~N(40, 0.25) according to the CLT, but when I tried calculating z = (36.2 - 40)/√(0.25), the z-score was 7.6 and way too far from the z-score table. Any ideas??

Thanks!

 

[clamtrox]

I don't understand what you are doing here. What's the number 0.25 and where does it come from? Shouldn't the CLT distribution have a standard deviation of 5 too?

 

[Ray Vickson]

Hi there,

Not sure if this is the right place, but the problem states that a battery has a population mean of 40 hours and standard deviation of 5. Let X represent the mean lifetime of batteries in a simple random sample size of 100. What is P(X <= 36.7)?

I tried computing this with X~N(40, 0.25) according to the CLT, but when I tried calculating z = (36.2 - 40)/√(0.25), the z-score was 7.6 and way too far from the z-score table. Any ideas??

Thanks!

The variance of X_bar (the mean of the X_i) is (1/100^2)*100*Var(X) = 5^2/100 as you implied, and that gives z = (36.7 - 40)/0.5 = -6.6 (not +7.6 as you wrote). This is still way beyond standard tables, but you can use instead the asymptotic expansion of the right-tail normal cdf. Let G(z) = P{Z > z} for Z ~ N(0,1). Express G as an integral, then integrate by parts. Letting
\phi(t) = \frac{1}{\sqrt{2 \pi}} e^{-t^2/2}, we have
G(z) = \int_z^{\infty} \phi(t) \, dt = \int_z^{\infty} \frac{1}{t} \cdot t \phi(t) ,\ dt <br /> = \frac{1}{z} \phi(z) - \int_z^{\infty} \frac{1}{t^2} \phi(t) \, dt.
Note that the "remainder" term ∫ phi/t^2 is less than---usually much less than-- 1/z^2 times the first term, so for z > 3 we get reasonable accuracy by just keeping the first term:
G(z) \approx \frac{1}{z} \phi(z) for z greater than tabulated values. The approximation can be improved by once again integrating by parts, writing phi/t^2 as (1/t^3)*(t \phi), etc. (However, this gives an _asymptotic_series, so after a certain number of terms the error starts to grow instead of diminish, which puts a limit on the attainable accuracy.) Since P{Z ≤ -6.6} = G(+6.6), the simple approximation gives P{Z ≤ -6.6} ≈ 0.21010e(-10) instead of the "exact" value of 0.20558e(-10).

BTW: the above expansion is well-known, and can be found in many sources.

RGV