We wish to estimate (with 99% confidence) the average amount of time students spend on homework per week. We survey a random sample of students and we get the following: µ = 10 0 = 0.2 n = 19 The data are not normally distributed. Which is the appropriate decision? You should use the normal distribution (zinterval). You should use student's t distribution (tinterval). Neither the normal distribution or the student's t distribution is an appropriate choice. You could use bootstrapping. We randomly sample 200 people in a specific town. Of these, 102 had dogs. Based on this, construct a 99 % confidence interval for the true population proportion of people in this town with dogs
Continuous Probability Distributions
Probability distributions are of two types, which are continuous probability distributions and discrete probability distributions. A continuous probability distribution contains an infinite number of values. For example, if time is infinite: you could count from 0 to a trillion seconds, billion seconds, so on indefinitely. A discrete probability distribution consists of only a countable set of possible values.
Normal Distribution
Suppose we had to design a bathroom weighing scale, how would we decide what should be the range of the weighing machine? Would we take the highest recorded human weight in history and use that as the upper limit for our weighing scale? This may not be a great idea as the sensitivity of the scale would get reduced if the range is too large. At the same time, if we keep the upper limit too low, it may not be usable for a large percentage of the population!
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