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Q. Four observations were made independently and at random from a normal distribution with mean of 80 and standard deviation of 20. Let X1 symbolize the value of the first observation, X2 the second, and so on through X4. A linear combination of these observed values creates a new variable Y such that
Y = 0.25X1 + 0. 25X2 + 0.25X3 + 0.25X4
Describe the sampling distribution of Y, along with its mean and variance. Within what two possible values of Y should the middle 99 percent of all sample values fall? The middle 50 percent?
A. The distribution of Y will depend on the distributions of the random variables of which it is a weighted sum. Because X1 to X4 are each normally distributed and statistically independent, and Y is a linear combination of these random variables, Y will also be normally distributed. In addition the mean of Y will be:
mean of Y = .25(mean of X1) + .25(mean of X2) + .25(mean of X3) + .25(mean of X4)
mean of Y = .25 * 80 + .25 * 80 + .25 * 80 + .25 * 80
mean of Y = 20 + 20 + 20 + 20 = 80.
The variance of Y will be:
variance of Y = (.25)^2(var of X1) + (.25)^2(var of X2) + (.25)^2(var of X3) + (.25)^2(var of X4).
variance of Y = .0625 * 20^2 + .0625 * 20^2 + .0625 * 20^2 + .0625 * 20^2
variance of Y = 25 + 25 + 25 + 25 = 100.
*** Don't forget to square the standard deviation to get the variance. ***
[Note that Y is the (simple, i.e., equally weighted) average of the X's. The X's are independent and identically distributed (i.e., they have the same mean and variance). In this case, the mean of Y is equal to the common mean of each random variable. This property has nothing to do with whether the X's are normally distributed. The varinace of Y is then the common variance divided by the number of samples (i.e., X's). In this case, the common variance is 20^2 or 400. Dividing 400 by 4 samples yields 100.]
Thus, Y is normally distributed with a mean of 80 and a variance of 100. The figure below shows the distribution of Y with the scale shown at the top. The bottom scale shows the standard score or z scale. (The z values are for a unit standard normal distribution, i.e., a normal distribution with a mean of zero and variance of 1.)
To find the two Y values that have 99% of the probability under the curve between them, make the following drawing:
We need to find the values for z at the dotted lines. If 99% are in the middle then 1% lies outside. If the 99% is centered, then the amount of probability below the left dotted line must be the same as the probability above the right dotted line, or 1%/2 or 1/2 of 1% or 0.5% or 0.005 as a fraction. Most normal curve tables show the amount of probability below a given z value and often start at a z of zero with a probability of 0.5. The probability below the right dotted line is 1 - 0.005 = 0.995. Look in the normal curve table in the probability column for the probability closest to 0.995. This z value is about 2.58 (in a table with z values rounded to two decimal places, the closest probability is 0.9950600). Thus, the P[ -2.58 < z < 2.58 ] = 0.99, approximately. Similarly, we want to find z1 and z2, such that:
P[ z1 < z < z2 ] = 50%,
such that z1 and z2 are symmetrically placed (i.e., z1 and z2 have the same magnitude and differ only in sign: abs(z1)=abs(z2), where "abs" represents the "absolute value"). z1 is at the left dotted line and z2 is at the right dotted line in the figure below.
If 50% is in between z1 and z2, then because the normal curve is symmetric, 25% must be below z1 (P[z < z1]=25%) and 25% above z2 (P[z > z2] = 25%. This implies that there must be 75% of the probability below z2 because P[z < z2] = 1 - P[z > z2] = 1 - 0.25 = 0.75 or 75%. The probability of 0.7485711 in the normal table corresponds to z2=0.67. Therefore, z1 must equal -0.67.
Finally, the standard score z values have to be converted from a normal distribution with mean of 0 and variance of 1 to a normal distribution with mean of 80 and variance of 100 (or standard deviation of sqrt(100)=10).
By definition, z = (Y - u)/s where u is the mean of Y and s is the standard deviation of Y. Solving for Y yields Y = u + sz. Using this relation to convert all the z values to Y values:
z: 2.58 Y = 80 + 10*2.58 = 105.8
z: -2.58 Y = 80 + 10*(-2.58) = 54.2
z: 0.67 Y = 80 + 10*0.67 = 86.7
z: -0.67 Y = 80 + 10*(-0.67) = 73.3
Thus, P[54.2 < Y < 105.8] = 0.99 and P[73.3 < Y < 86.7] = 0.50.
This page was last updated on October 5, 1997 and has been accessed 371 times.
Copyright ©1997 Richard A. Bilonick