Let X be a normal random variable with mean = 10 and standard deviation - 5. Let
ID: 3156669 • Letter: L
Question
Let X be a normal random variable with mean = 10 and standard deviation - 5. Let X be the sample average of five observations of X. What is the probability distribution of X? Let Z = (x -10)/5 and let Y = (Z_1)^2 + (Z_2)^2 + (Z_3)^2 + (Z_4)^2 + (Z_5)^2 where the Z_i are independent observations of Z. What is the probability distribution of Y? If you conduct a "t-test" or more specifically compute the test statistic t_0 = x - mu_c/s/squareroot n from n observations of a random variable X for the purpose of testing the hypothesis that the mean of X = mu_0, why should you check that that a normal distribution is a reasonable model for X? Let X be a random variable with mean = 10 and standard deviation = 5 (normality is not specified). If Y is the sum of 20 independent observations of X. What is the approximate distribution of Y? What is the approximate distribution of Y = Y/20? What is the mean and standard deviation of Y?Explanation / Answer
1. (a) X be a normal random variable with mean = 10 and Standard deviation = 5
X' is the average of 5 obervation of X so E(X) = µ ( since mean is the unbiased estimate of the population mean
So, E(X) = = X' = 10 and Standard Deviation of X' = / n = 5 / 5 = 5
So the distribution X' will follow the standard normal distribution with the variate Z = (X' - 10) / 5
(b) Z = (X - 10) /5 is a standard normal variate, Y is the sum of squares of 5 independent standard normal variates. So, Y is Chi-square Probability distribution
2. For the purpose of testing the mean = µ0 , the test statistic t = (x' - µ0 ) / s/n follows t-distribution with n-1 df
if t < 30, otherwise it follows normal distribution.
3. a. The distribution of Y is normal
b. Y/20 is also normal
c. Mean = 10 and Standard deviation = 5/(20) =5/4.472136 = 1.118.
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