Suppose x has a distribution with μ = 30 and σ = 26.   (a) If a random sample of size n = 31 is drawn, find μx, σ x and P(30 ≤ x ≤ 32). (Round σx to two decimal places and the probability to four decimal places.) μx =  σ x =  P(30 ≤ x ≤ 32) =  (b) If a random sample of size n = 65 is drawn, find μx, σ x and P(30 ≤ x ≤ 32). (Round σ x to two decimal places and the probability to four decimal places.) μx =  σ x =  P(30 ≤ x ≤ 32) =  (c) Why should you expect the probability of part (b) to be higher than that of part (a)? (Hint: Consider the standard deviations in parts (a) and (b).) The standard deviation of part (b) is  ---Select--- larger than the same as smaller than part (a) because of the  ---Select--- same larger smaller sample size. Therefore, the distribution about μx is  ---Select--- wider the same narrower .

Suppose that X has a Weibull distribution with β = 2 and δ = 2400. Determine the following. a. P(X > 5000) = b. For an exponential random variable with the same mean as the Weibull distribution  P(X > 5000) =

If we let X1, X2,...X10 be a random sample from the distribution N( μ, σ), then what is the value k so that: P( (X-μ)/(σ/√10) < k ) = .05 P( (X-μ)/(S/√10) < k ) = .05 P( k < (X-μ)/(S/√10) ) = .05

The third worksheet labeled sample B is a simple random sample with replacement, with seven observations in the sample, from some population. The index i in the first column is just an integer name for each observation. The specific values x in the second column are measures of a random variable X distributed in the population. Now suppose this random variable X is known to have a Normal distribution in the sampled population. You still do not know the population mean and must estimate it from the sample, as you did earlier. BUT NOW, in this question, you know in advance that the true population standard deviation of the random variable X is 4.  Compute the standard error (of the sample mean), given this new knowledge situation. i xi 1 74.2 2 72.79 3 70.35 4 74.37 5 73.34 6 78.96 7 77.99