In the early 1800s, astronomers measuring the same star got different answers every time. Telescopes weren't perfect. The atmosphere interfered.

Decades later, Adolphe Quetelet applied Gauss's curve to people: heights, weights, chest measurements, waist measurements, blood pressure, reaction times, etc…

Consider a data set consisting of vehicle emissions. We can visualize these data with a relative frequency histogram, with class intervals chosen so each rectangle represents a reasonably large number of vehicles.

Vehicle emissions is a continuous random variable.

Suppose the sample was very large, consisting of millions. Because emissions is continuous, there would be plenty of vehicles for each rectangle even with small class intervals.

Such a histogram would look smooth and could be approximated by a curve. The curve used to describe the distribution of this variable is called a probability density curve. The probability density curve tells what proportion of the data falls within a given interval.

The area under a probability density curve between two values $a$ and $b$ represents the proportion of a population whose values are between $a$ and $b$. It can also be interpreted as the probability that a randomly selected value from the population is between $a$ and $b$.

The region above a single point has no width, thus no area. Therefore, if $X$ is a continuous random variable, $P(X=a)=0$ for any number $a$.

This means that $P(a<X<b)=P(a\le X\le b)$ for any numbers $a$ and $b$.

For any probability density curve, the area under the entire curve is 1, because this area represents the entire population.

Before we tackle the normal curve, let's start with the simplest probability density curve: one where every value in an interval is equally likely.

A uniform distribution is one in which values in any region are equally likely. The probability density curve for a uniform distribution is a horizontal line.

Imagine that when a traffic light turns red, it stays red for 30 seconds before turning green. If you pull up to the light when it is red, the amount of time you will wait before the light turns green is uniformly distributed between 0 and 30 seconds.

Find the probability that the waiting time is between 5 and 15 seconds.

Probability density curves come in many varieties, depending on the characteristics of the populations they represent. Many important statistical procedures can be carried out using only one type of probability density curve, called a normal curve.

Normal distributions have one mode and the distributions are symmetric around the mode. Normal curves extend infinitely far both to the right and to the left.

The mean determines the location of the peak. The standard deviation measures spread. Large values produce a wide, flat curve; small values produce a tall, narrow one.

In a normal distribution, the mean, median, and mode are all equal.

The area under a normal curve represents a proportion of the population or can be interpreted as a probability. It is therefore necessary to find areas under the normal curve other than those specified by the Empirical Rule.

The $z$-score of a data value represents the number of standard deviations that data value is above or below the mean.

If $x$ is a value from a normal distribution with mean $\mu$ and standard deviation $\sigma$, we can convert $x$ to a $z$-score by using a method known as standardization.

The color temperature of LED bulbs from one production line is normally distributed with mean $\mu =3000$ K and standard deviation $\sigma =200$ K. Find the $z$-score of a bulb measured at $x=2920$ K.

On the TI-84 Plus, the normalcdf command finds areas under a normal curve.

It takes four inputs: lower bound, upper bound, mean, and standard deviation.

Access it by pressing 2nd → VARS.

Find the area to the left of $z=1.26$.

The area we want is on the left, so choose Area to the left of x.

The value is a $z$-score, so the mean is 0 and the standard deviation is 1.

Use normalcdf with −1E99 (or some really small number) as the lower endpoint, 1.26 as the upper endpoint, 0 as the mean, and 1 as the standard deviation.

A study reported that the length of pregnancy is approximately normally distributed with mean $\mu =272$ days and standard deviation $\sigma =9$ days. What proportion of pregnancies last less than 259 days?

The area we want is on the left, so choose Area to the left of x.

We use normalcdf(−1E99,259,272,9).

A study reported that the length of pregnancy is approximately normally distributed with mean $\mu =272$ days and standard deviation $\sigma =9$ days. What proportion of pregnancies last longer than 280 days?

The area we want is on the right, so choose Area to the right of x.

We use normalcdf(280,1E99,272,9).

A study reported that the length of pregnancy is approximately normally distributed with mean $\mu =272$ days and standard deviation $\sigma =9$ days. A pregnancy is considered full-term if it lasts between 252 days and 298 days. What proportion of pregnancies are full-term?

The area we want falls between two values, so choose Area between two values.

We use normalcdf(252,298,272,9).

a) The number of hours a company's employees work remotely each week is normally distributed with mean $\mu =24$ and standard deviation $\sigma =6$. Find the probability that an employee works fewer than 12 hours remotely.

b) Waiting times in a clinic are normally distributed with mean $\mu =40$ minutes and standard deviation $\sigma =10$ minutes. Find the probability that a patient waits more than 53 minutes.

c) Temperature fluctuations in a laboratory are normally distributed with mean $\mu =5.6$ degrees and standard deviation $\sigma =4$ degrees. Find the probability that a fluctuation is between $-1.6$ and 8 degrees.

The invNorm command returns the value from a normal population with a given area to its left.

Access via 2nd → VARS (DISTR)

Scores on an IQ test are normally distributed with mean $\mu =100$ and standard deviation $\sigma =15$. What score separates the upper 2% of IQ scores from the lower 98%?

The area 0.02 is on the right, so choose x with area to its right.

The area 0.02 is on the right, so the area on the left is 0.98.

Find the IQ scores that separate the middle 90% of the scores from the top and bottom 5%.

$\mu =100$

$\sigma =15$

The middle 90% falls between two cutoffs, so choose Middle % boundaries.

The middle area is 0.90. That is a proportion, not 90.

$\mu =100$

$\sigma =15$

For the lower cutoff, area on left = 0.05. For the upper cutoff, area on left = 0.95.

You have two ways to get to campus. Both take 20 minutes on average. Route A is almost always close to it. Route B is all over the place.

We’ll model both as normal: Route A with $\mu =20$ and $\sigma =2.5$ minutes, Route B with $\mu =20$ and $\sigma =5$.