5.3 Normal Distribution (Lap Times)

Aligned outcomes:

Learning Objectives

By the end of this section, you will be able to:

In this section, you will learn to:

collect a stratified sample of lap times and organize it into a histogram;
estimate the theoretical normal distribution that best matches the sample you collected;
calculate the median, quartiles, IQR, percentiles, and a tail probability from your own data;
calculate those same quantities from the theoretical normal model and compare the two sets of answers;
decide whether the normal model is a good description of the data you collected, and defend that decision.

This section is a lab, not a reading. In 6.1 and 6.2 you were handed a mean and a standard deviation and asked to compute probabilities from them. Here nobody hands you anything. You collect real lap times from a real motorcycle racer, work out what normal distribution they look like they came from, and then check whether that model actually earns its keep.

The whole lab turns on one comparison, so it is worth naming up front. You are going to answer the same five questions twice — once by counting your own data, and once by trusting a smooth curve — and then look at how far apart the two answers landed.

Class Time:

Names:

5.3.1 Stats Lab: Normal Distribution (Lap Times)

Definition 5.3.1: Empirical Distribution

The empirical distribution of a variable is the distribution you get by counting the data you actually collected: the frequency of each value or interval, the quartiles you find by ordering the list, the percentiles you read off positions in that list. It has no formula behind it. Every number in it comes from a value someone wrote down.

Definition 5.3.2: Theoretical Distribution

The theoretical distribution is a named probability model — here a normal distribution $X \sim N(\mu, \sigma)$ — chosen to stand in for the process that generated the data. Its median, quartiles, percentiles, and probabilities come from the model's formula, not from any particular data set, so the same model gives the same answers no matter which sample you drew.

A tracing over a photograph

The empirical distribution is the photograph — every bump and gap exactly as your sample happened to fall. The theoretical distribution is the smooth curve someone traced over it. The tracing is easier to work with and it lets you answer questions your photograph never covered, but it is only worth using if it followed the outline closely in the first place. This lab makes you check the tracing against the photograph.

Both objects answer the question "what fraction of lap times are below 130 seconds?" — and they will not give you the same number. Neither is cheating. The empirical answer is a fact about the 36 times you sampled and nothing more; ask a different classmate with a different sample and you get a different fact. The theoretical answer is a claim about every lap the racer will ever run, and it is only as good as the assumption that lap times are normally distributed. When the two land close together, that assumption is doing honest work. When they land far apart, something about the normal model does not fit — maybe the times are skewed by a handful of very slow laps, maybe your sample was too small for its quartiles to settle down. Either way, the gap between the two answers is the finding, not an error to be explained away.

Try It Now 5.3.1

Hannah sampled 36 lap times and found that 9 of them came in slower than 131 seconds. Her theoretical model, $N(129.25, 2.50)$, gives $P(X > 131) = 0.2420$. Which of these two numbers is the empirical probability and which is the theoretical one, and why are they not equal?

Solution

Step 1 — identify the empirical one. It is the one that came from counting. Nine of Hannah's 36 sampled times were slower than 131 seconds, so

$$ P(X > 131) = \frac{9}{36} = 0.2500 $$

That 0.2500 is a fact about the 36 values on her worksheet.

Step 2 — identify the theoretical one. The 0.2420 came out of the normal model, by converting 131 to a $z$-score and reading the area to its right. No lap time was consulted to produce it.

Step 3 — say why they differ. They are answers to different questions. The empirical number describes one sample of 36; draw a different 36 and it will move. The theoretical number describes every lap the model claims she will ever run, and it will not move at all unless you change $\mu$ or $\sigma$.

Answer: 0.2500 is empirical, 0.2420 is theoretical. They land within a hundredth of each other, which is good news for the normal model — a sample of 36 agreeing that closely with the curve is evidence the curve is describing the process honestly.

Directions

Round the relative frequencies and probabilities to four decimal places. Carry all other decimal answers to two places.

5.3.2 Collect the Data

Use the data from Appendix C. Use a stratified sampling method by lap (races 1 to 20) and a random number generator to pick six lap times from each stratum. Record the lap times below for laps two to seven.

**Table 5.3.1** — Six randomly selected lap times from each of laps two through seven (36 values total).
Lap 2	Lap 3	Lap 4	Lap 5	Lap 6	Lap 7

Try it in rāSHio

Once the grid above is full, open rāSHio and choose File → Delimited List… to get all 36 of your lap times into the spreadsheet in one paste. Commas, semicolons, tabs, and one-per-line all parse, so you can type the values straight off your paper grid column by column without reformatting anything first.

Figure 5.3.1 — Getting your 36 collected lap times into rāSHio: File → Delimited List…

The strata here are the laps, not the races. Every lap number gets six of its times into your sample, so a lap the racer consistently runs slowly cannot be crowded out by chance. That is the point of stratifying: it guarantees coverage you would otherwise be hoping for. Use the random number generator to pick which six races supply lap 2's times, then do it again for lap 3, and so on — the randomness lives inside each lap, and the even split across laps is imposed by the design.

Try it in rāSHio

You need a random number generator to decide WHICH six races supply each lap's times. Open rāSHio, choose File → Random Numbers…, and set Min 1, Max 20 (the twenty races) and How many 6 — then repeat it once per lap, so laps two through seven each get their own independent draw.

Figure 5.3.2 — Drawing which six races supply each lap's times: File → Random Numbers…

Construct a histogram. Make five to six intervals. Sketch the graph using a ruler and pencil. Scale the axes.

Figure 5.3.3 — Blank axes for sketching your lap-time histogram, with lap time on the horizontal axis and frequency on the vertical axis.

Figure 5.3.3 — Blank axes for sketching your lap-time histogram: lap time across the bottom, frequency up the side.

Before you draw a single bar, decide on your intervals and write them along the horizontal axis. Five or six intervals spread across your smallest and largest lap time is the target. Too few and every value piles into two fat bars that show you nothing; too many and each bar holds one or two values, so the shape you see is mostly noise. Scale the vertical axis after you have counted the tallest interval, not before.

Try it in rāSHio

Sketch the histogram by hand first — that is the part of the lab that teaches you what an interval choice does. Then open rāSHio and choose Graph → Histogram on your 36 lap times to check yourself: set the bin start and width to match the intervals you drew, and the bars it returns should be the ones you counted.

Calculate the following:

a. $\bar{x} = $ ____

b. $s = $ ____

Draw a smooth curve through the tops of the bars of the histogram. Write one to two complete sentences to describe the general shape of the curve. Keep it simple: does the graph go straight across, does it have a v-shape, does it have a hump in the middle or at either end?

Try It Now 5.3.2

Wen and their wife Mei split the stopwatch duty for one stratum. The six lap times they recorded came out as 127.6, 129.1, 130.4, 128.2, 131.0, and 129.7 seconds. Calculate the sample mean and the sample standard deviation, carrying your answers to two decimal places.

Solution

Step 1 — add the six values and divide by six.

$$ \bar{x} = \frac{127.6 + 129.1 + 130.4 + 128.2 + 131.0 + 129.7}{6} = \frac{776.0}{6} = 129.33 $$

Step 2 — find how far each value sits from the mean, and square each distance. Working from the unrounded mean of 129.3333:

$$ (-1.7333)^2 = 3.0044 \qquad (-0.2333)^2 = 0.0544 \qquad (1.0667)^2 = 1.1378 $$ $$ (-1.1333)^2 = 1.2844 \qquad (1.6667)^2 = 2.7778 \qquad (0.3667)^2 = 0.1344 $$

Step 3 — add the squares, divide by $n - 1$, take the square root. We divide by 5 rather than 6 because this is a sample standard deviation, not a population one.

$$ s = \sqrt{\frac{8.3933}{5}} = \sqrt{1.6787} = 1.30 $$

Answer: $\bar{x} = 129.33$ seconds and $s = 1.30$ seconds. Notice how small that standard deviation is next to the mean — a spread of barely over a second around a lap that takes more than two minutes. A racer that consistent is exactly the kind of process a normal model has a chance of describing well.

Try it in rāSHio

With your 36 lap times loaded, rāSHio will do the arithmetic of the Try It Now above on the whole sample at once: choose Stats → Summary Statistics and read the sample mean and sample standard deviation straight off the panel. Work $\bar{x}$ and $s$ out by hand first, then use this to check them.

Figure 5.3.4 — Reading the sample mean and standard deviation in rāSHio: Stats → Summary Statistics.

5.3.3 Analyze the Distribution

Using your sample mean, sample standard deviation, and histogram to help, what is the approximate theoretical distribution of the data?

$X \sim$ ____(____, ____)
How does the histogram help you arrive at the approximate distribution?

You are picking a model here, and the histogram is your evidence. A single hump near the middle with the bars falling away at roughly the same rate on both sides is what a normal distribution looks like when you only have three dozen values; that shape is your licence to write $N$ rather than some other letter. The two blanks inside the parentheses come straight from step 3 above — the sample mean stands in for $\mu$ and the sample standard deviation stands in for $\sigma$. They are estimates, not the real thing, which is exactly why the rest of the lab makes you check them.

Try It Now 5.3.3

Ryan's 36 lap times gave $\bar{x} = 129.33$ seconds and $s = 1.30$ seconds. His histogram has one clear hump near 129 seconds, with the bars shrinking at about the same rate on the left and the right. Write the approximate theoretical distribution, and say what in the histogram justifies each part of your answer.

Solution

Step 1 — name the family from the shape. One hump, roughly symmetric, tapering on both sides. That is the normal family, so the model is $N$.

Step 2 — estimate the center. The hump sits near 129 seconds, and the sample mean is 129.33. For a symmetric histogram the mean lands under the peak, so the two agree and $\mu \approx 129.33$.

Step 3 — estimate the spread. The sample standard deviation is 1.30 seconds, so $\sigma \approx 1.30$. You can sanity-check this against the histogram: about 95% of a normal distribution falls within two standard deviations of the center, so nearly all the bars should sit between $129.33 - 2(1.30) = 126.73$ and $129.33 + 2(1.30) = 131.93$.

Answer: $X \sim N(129.33, 1.30)$. The single symmetric hump justifies the normal family, the location of the peak justifies the mean, and the width of the bars around that peak justifies the standard deviation. If the histogram had a long tail stretching to the right — a few laps far slower than the rest — none of this would follow, and the honest answer would be that the normal model does not fit.

5.3.4 Describe the Data

Use the data you collected to complete the following statements.

The IQR goes from ____ to ____.
$IQR = $ ____. $(IQR = Q_3 - Q_1)$
The $15^{\text{th}}$ percentile is ____.
The $85^{\text{th}}$ percentile is ____.
The median is ____.
The empirical probability that a randomly chosen lap time is more than 130 seconds is ____.
Explain the meaning of the $85^{\text{th}}$ percentile of this data.

Every one of these comes from counting, not from a formula about $\mu$ and $\sigma$. Order your 36 times smallest to largest and work down the list: the median splits it in half, $Q_1$ is the median of the bottom half and $Q_3$ the median of the top half, and a percentile is found by the index rule $i = \frac{k}{100}(n + 1)$ from Section 2.3. The empirical probability is the plainest one of all — count how many of your times exceeded 130 seconds and divide by 36.

Try It Now 5.3.4

Layla and her wife Noor pooled a smaller sample of twelve lap times, which they have already ordered for you:

125.9; 126.4; 127.1; 127.8; 128.3; 128.9; 129.4; 130.2; 130.8; 131.5; 132.6; 134.1

Find the median, $Q_1$, $Q_3$, the IQR, the 15th and 85th percentiles, and the empirical probability that a lap time is more than 130 seconds.

Solution

Step 1 — the median. With $n = 12$ the middle falls between the 6th and 7th values.

$$ \text{median} = \frac{128.9 + 129.4}{2} = 129.15 $$

Step 2 — the quartiles. The bottom half is the first six values, so $Q_1$ is the average of the 3rd and 4th. The top half is the last six, so $Q_3$ is the average of the 9th and 10th.

$$ Q_1 = \frac{127.1 + 127.8}{2} = 127.45 \qquad Q_3 = \frac{130.8 + 131.5}{2} = 131.15 $$ $$ IQR = 131.15 - 127.45 = 3.70 $$

Step 3 — the percentiles, using the index rule. For the 15th percentile, $i = \frac{15}{100}(12 + 1) = 1.95$. That is not a whole number, so round it down to 1 and up to 2 and average those two data values.

$$ P_{15} = \frac{125.9 + 126.4}{2} = 126.15 $$

For the 85th percentile, $i = \frac{85}{100}(13) = 11.05$, so average the 11th and 12th values.

$$ P_{85} = \frac{132.6 + 134.1}{2} = 133.35 $$

Step 4 — the empirical probability. Five of the twelve times are above 130 seconds: 130.2, 130.8, 131.5, 132.6, and 134.1.

$$ P(X > 130) = \frac{5}{12} = 0.4167 $$

Answer: median 129.15, $Q_1 = 127.45$, $Q_3 = 131.15$, $IQR = 3.70$, $P_{15} = 126.15$, $P_{85} = 133.35$, and an empirical probability of 0.4167. The 85th percentile says that 85% of these lap times were faster than 133.35 seconds and 15% were slower — a lap over 133.35 seconds was one of her worst.

Try it in rāSHio

The empirical five-number summary is exactly what a box plot draws. Open rāSHio and choose Graph → Box Plot on your 36 lap times to see $Q_1$, the median, and $Q_3$ marked on the same axis, with the box itself spanning the IQR you just calculated by hand.

Figure 5.3.5 — Seeing your empirical quartiles and IQR as a picture: Graph → Box Plot.

5.3.5 Theoretical Distribution

Using the theoretical distribution, complete the following statements. You should use a normal approximation based on your sample data.

The IQR goes from ____ to ____.
$IQR = $ ____.
The $15^{\text{th}}$ percentile is ____.
The $85^{\text{th}}$ percentile is ____.
The median is ____.
The probability that a randomly chosen lap time is more than 130 seconds is ____.
Explain the meaning of the $85^{\text{th}}$ percentile of this distribution.

Same questions, different machinery

These are word-for-word the questions you just answered by counting, and that repetition is deliberate. This time your ordered list of lap times never gets opened. Every answer comes out of $N(\mu, \sigma)$ and a $z$-score, which means you could answer them for a lap the racer has not run yet.

Two facts about the normal distribution do most of the work here. Its median equals its mean, because the curve is symmetric about $\mu$ and half the area sits on each side. And every percentile is the mean plus some fixed number of standard deviations: $x = \mu + z\sigma$, where $z$ is whatever score cuts off the area you want. The quartiles are just the 25th and 75th percentiles, so $Q_1 = \mu - 0.6745\sigma$ and $Q_3 = \mu + 0.6745\sigma$, which makes the theoretical IQR exactly $1.349\sigma$ — the same multiple of $\sigma$ for every normal distribution there has ever been. That last fact is worth sitting with, because it is the sharpest illustration of what a theoretical distribution buys you and what it costs. Once you commit to the normal model, the spacing between the quartiles is no longer something your data gets to decide; it is fixed by the model, and your data only gets a vote on $\sigma$. If your empirical IQR comes out at 3.70 while the model insists on 3.37, the model has not made an arithmetic mistake — it has told you that your data's middle half is wider than a normal distribution with that standard deviation would produce. Going the other direction, a probability like $P(X > 130)$ is where the theoretical model genuinely outperforms counting: your sample might contain no laps at all above 134 seconds, so the empirical probability of one is flatly zero, while the model still assigns it a small positive chance. Neither answer is wrong. They are answers to different questions.

Try It Now 5.3.5

Diego is working from the theoretical model $X \sim N(129.25, 2.50)$. Help him find the median, $Q_1$, $Q_3$, the IQR, the 15th and 85th percentiles, and $P(X > 130)$. Round the probability to four decimal places and everything else to two.

Solution

Step 1 — the median. A normal curve is symmetric about its mean, so the median is the mean: 129.25 seconds.

Step 2 — the quartiles. The 25th and 75th percentiles sit at $z = -0.6745$ and $z = +0.6745$.

$$ Q_1 = 129.25 - 0.6745(2.50) = 127.56 \qquad Q_3 = 129.25 + 0.6745(2.50) = 130.94 $$ $$ IQR = 130.94 - 127.56 = 3.37 $$

Step 3 — the 15th and 85th percentiles. The $z$-score with 15% of the area below it is $-1.0364$, and by symmetry the 85th percentile sits at $+1.0364$.

$$ P_{15} = 129.25 - 1.0364(2.50) = 126.66 \qquad P_{85} = 129.25 + 1.0364(2.50) = 131.84 $$

Step 4 — the probability of a lap slower than 130 seconds. Convert 130 to a $z$-score, then take the area to its right.

$$ z = \frac{130 - 129.25}{2.50} = 0.30 $$ $$ P(X > 130) = P(Z > 0.30) = 1 - 0.6179 = 0.3821 $$

Answer: median 129.25, $Q_1 = 127.56$, $Q_3 = 130.94$, $IQR = 3.37$, $P_{15} = 126.66$, $P_{85} = 131.84$, and $P(X > 130) = 0.3821$. The 85th percentile tells Diego the model predicts 85% of the racer's laps come in under 131.84 seconds. Note that this is a claim about every lap she will ever run, not about the ones he happened to sample.

Try it in rāSHio

Every theoretical answer in this section is an area or a cut-point on one normal curve, and rāSHio will give you both without a table. Choose Distributions → Normal, enter your own $\mu$ and $\sigma$, then ask it for the area to the right of 130 seconds and for the values at the 15th and 85th percentiles.

Figure 5.3.6 — Computing the theoretical percentiles and P(X > 130) in rāSHio: Distributions → Normal.

5.3.6 Discussion Questions

Do the data from the section titled Collect the Data give a close approximation to the theoretical distribution in the section titled Analyze the Distribution? In complete sentences and comparing the result in the sections titled Describe the Data and Theoretical Distribution, explain why or why not.

Answer this by lining your two sets of numbers up side by side — empirical median against theoretical median, empirical IQR against theoretical IQR, and so on down the list — and saying how far apart each pair landed. A gap of a tenth of a second is agreement; a gap of three seconds is not. Point at the specific comparison that convinced you rather than giving a general impression, and if one quantity agreed while another did not, say which and offer a reason.

Key Terms

empirical distribution — the distribution obtained by counting the data you actually collected; its quartiles, percentiles, and probabilities all come from positions in the ordered sample.

theoretical distribution — a named probability model such as $N(\mu, \sigma)$ used to stand in for the process that generated the data; its quantities come from the model's formula rather than from any one sample.