What is the probability that in the preceding example X exceeds 12?

The probability that X exceeds 12 is the same as that Z exceeds 2. From Table D.1 it is obvious that this probability is (0.5 - 0.4772) or 0.0228.

4. Let X1 — N(|1, a^) and X2 — N(|2, a22) and assume that they are independent. Now consider the linear combination

where a and b are constants. Then it can be shown that

This result, which states that a linear combination of normally distributed variables is itself normally distributed, can be easily generalized to a linear combination of more than two normally distributed variables.


5. Central limit theorem. Let X1, X2,..., Xn denote n independent random variables, all of which have the same PDF with mean = p and variance = o2. Let X=ZXi/n (i.e., the sample mean). Then as n increases indefinitely (i.e., n ^ to),

That is, X approaches the normal distribution with mean p and variance o2 /n. Notice that this result holds true regardless of the form of the PDF. As a result, it follows that

That is, Z is a standardized normal variable.

6. The third and fourth moments of the normal distribution around the mean value are as follows:

Third moment: E(X — p)3 = 0 Fourth moment: E(X — p)4 = 3o4

Note: All odd-powered moments about the mean value of a normally distributed variable are zero.

7. As a result, and following the measures of skewness and kurtosis discussed earlier, for a normal PDF skewness = 0 and kurtosis = 3; that is, a normal distribution is symmetric and mesokurtic. Therefore, a simple test of normality is to find out whether the computed values of skewness and kurtosis depart from the norms of 0 and 3. This is in fact the logic underlying the Jarque-Bera (JB) test of normality discussed in the text:

where S stands for skewness and K for kurtosis. Under the null hypothesis of normality, JB is distributed as a chi-square statistic with 2df.

8. The mean and the variance of a normally distributed random variable are independent in that one is not a function of the other.

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