Computing probabilities with the CDF

Hey guys,

could someone help me with this question:

A normally distributed random variable has a mean of 100 and a standard deviation of 12. The probability of observing a value greater than 82 is the cumulative distribution function (cdf) of the standard normal variable: 1 – N(1.5). N(–1.5). N(1.5).

Answer: The standardized value of this normal distribution can be obtained using the formula =(X-μ)/σ =(82-100)/12=- 1.5. The cdf of N(–1.5) provides the probability of a value less than or equal to 82.

The bolded sentence seems to be a good indication that they appear to be answering a different question. Here is my solution.

In order to compute the probability observing a value greater than 82, we can first compute the probability of a value smaller than 82, (because of the symmetriy of the normal distribution this is the same as a value greater than 82) and then subtract that from 1.So:

Probability of a value smaller than 82= N(-1.5) b/c if symmetry this is identical to 1-N(1.5)

Now we want to know the probability of observing a value greater than 82, which would then be:

Probability of a value bigger than 82= 1-N(-1.5) b/c if symmetry this is identical to 1-(1-N(1.5))=N(1.5)

Is this correct?

The answer should be N(1.5).

P(X > 82) = P(X < 118) = P(X ≤ 118) = P(Z ≤ 1.5)

That is what I thought, thank you so much S2000!!

My pleasure.