Consider any random observation of some process or population. Let this “random observation” be modeled by a normally distributed random variable X with mean μ, and variance σ2. Note that now, in contrast to the previous problem, we don't know the values of μ, and σ. We'll get around this by taking a random sample from the distribution of X.(a) What's the underlying random experiment for X? (b) What was “the model” when we were generating a confidence interval for an unknown mean? (c) For a prediction interval, the model is X – . Show that this model is normally distributed with mean zero and with variance given by(d) For part (c), show that there's a 95% chance for X to take a value within 1.96 standard errors of  .(e) Show that for a prediction interval, the endpoints are with the standard error determined by the formula in part (c). Explain what to do if σ is unknown.(f) Suppose the previous ten cups from the drink machine had a mean of 6.6 ounces with a standard deviation of 0.27 ounces. How much drink should I anticipate when I drop my coins in the machine?(g) Now suppose you want a prediction interval for the mean of m independent future observations of X. Now the model is where A is the average of m observations. Show that the squared standard error of the model is(h) How much drink will three of us obtain from the machine given the information in part (f)(i) If n is large,  is approximately normally distributed even if X is not. That's the Central Limit Theorem. So why do we say our prediction interval is not valid unless X is normally distributed? Why couldn't we eliminate that assumption in the large sample case? Where have we used the normality of X in an essential way (even if n is large)?(j) The confidence coefficient, let's say 95%, for a confidence interval says that, on average, 95 of 100 intervals obtained by the given technique will contain the parameter. What's the precise interpretation of the confidence level for a prediction interval to predict one future observation?