3.5: Continuous Distributions
- Page ID
- 56916
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)So far in this chapter we’ve discussed cases where the outcome of a variable is discrete. In this section, we consider a context where the outcome is a continuous numerical variable.
Figure 3.24 shows a few different hollow histograms for the heights of US adults. How does changing the number of bins allow you to make different interpretations of the data?
Solution
Adding more bins provides greater detail. This sample is extremely large, which is why much smaller bins still work well. Usually we do not use so many bins with smaller sample sizes since small counts per bin mean the bin heights are very volatile.
What proportion of the sample is between cm and cm tall (about 5’11" to 6’1")?
Solution
We can add up the heights of the bins in the range cm and and divide by the sample size. For instance, this can be done with the two shaded bins shown in Figure 3.25.
The two bins in this region have counts of 195,307 and 156,239 people, resulting in the following estimate of the probability:
\[\begin{aligned} \frac{195307 + 156239}{\text{3,000,000}} = 0.1172\end{aligned}\]
This fraction is the same as the proportion of the histogram’s area that falls in the range to cm.
From histograms to continuous distributions
Examine the transition from a boxy hollow histogram in the top-left of Figure 3.24 to the much smoother plot in the lower-right. In this last plot, the bins are so slim that the hollow histogram is starting to resemble a smooth curve. This suggests the population height as a continuous numerical variable might best be explained by a curve that represents the outline of extremely slim bins.
This smooth curve represents a probability density function (also called a density or distribution), and such a curve is shown in Figure 3.26 overlaid on a histogram of the sample. A density has a special property: the total area under the density’s curve is 1.
Probabilities from continuous distributions
We computed the proportion of individuals with heights to cm in Example 3.72 as a fraction:
\[\begin{aligned} \dfrac{\text{number of people between 180 and 185}}{\text{total sample size}}\end{aligned}\]
We found the number of people with heights between 180 and 185 cm by determining the fraction of the histogram’s area in this region. Similarly, we can use the area in the shaded region under the curve to find a probability (with the help of a computer):
\[\begin{aligned} P(\text{height between 180 and 185}) = \text{area between 180 and 185} = 0.1157\end{aligned}\]
The probability that a randomly selected person is between 180 and 185 cm is 0.1157. This is very close to the estimate from Example 3.72: 0.1172.
Three US adults are randomly selected. The probability a single adult is between and cm is 0.1157.
- What is the probability that all three are between and cm tall?
- What is the probability that none are between and cm?
- Answer
-
Brief answers:
- \(0.1157 × 0.1157 × 0.1157 = 0.0015.\)
- \((1 − 0.1157)^3 = 0.692\)
What is the probability that a randomly selected person is exactly 180cm? Assume you can measure perfectly.
Solution
This probability is zero. A person might be close to 180 cm, but not exactly 180 cm tall. This also makes sense with the definition of probability as area; there is no area captured between 180 cm and 180 cm.
Suppose a person’s height is rounded to the nearest centimeter. Is there a chance that a random person’s measured height will be cm?
- Answer
-
This has positive probability. Anyone between 179.5 cm and 180.5 cm will have a measured height of 180 cm. This is probably a more realistic scenario to encounter in practice versus Example 3.74
- The 52 cards are split into four : \(\clubsuit\) (club), \(\diamondsuit\) (diamond), \(\heartsuit\) (heart), \(\spadesuit\) (spade). Each suit has its 13 cards labeled: , , ..., , (jack), (queen), (king), and (ace). Thus, each card is a unique combination of a suit and a label, e.g. and . The 12 cards represented by the jacks, queens, and kings are called . The cards that are \(\diamondsuit\) or \(\heartsuit\) are typically colored red while the other two suits are typically colored black.
- The actual proportion of the U.S. population that is is about 50%, and so we use 0.5 for the probability of sampling a woman. However, this probability does differ in other countries.
- The 10% guideline is a rule of thumb cutoff for when these considerations become more important.
- \(\mu = \int xf(x)dx\) where \(f(x)\) represents a function for the density curve.
- If \(X\) and \(Y\) are random variables, consider the following combinations: \(X^{1+Y}\), \(X\times Y\), \(X/Y\). In such cases, plugging in the average value for each random variable and computing the result will not generally lead to an accurate average value for the end result.