Probability Question about Normal Distribution?

瑋爺

2013-10-13 6:52 am

Given Y = (10^x) + 3 and X~N(2,1), what is Pr(Y>300) ?

My approach:

Pr(Y>300)
= 1 - Pr(Y<=300)
= 1- Pr((10^x)+3<=300)
= 1 - Pr( (10^x) <= 297)
.
.
.

Is my approach correct ?

As I am really confused by the lognormal distribution and the normal distribution.

thx

回答 (1)

az_lender

2013-10-15 4:36 am

✔ 最佳答案

The reason why nobody is answering is that nobody understands your notation "X~N(2,1)". Does this perhaps signify that X is a random variable, normally distributed with a mean of 2 and a standard deviation of 1??? I'm just inventing an interpretation!

The probability that Y > 300 is indeed the same as the probability that 10^X > 297, so it's the same as the probability that X > 2.47276. If X is a normally-distributed random variable with a mean of 2 and a standard deviation of 1, then the z-score associated with 2.47276 is 0.47276; the probability that X < 2.47276 is 0.6818 (look at a normal distribution table and interpolate between z=0.47 and z=0.48).

So the probability that X > 2.47276 is 0.3192, and that's the probability that Y > 300.

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