Chapter 6: Statistics: Normal Distribution
Objective
Here you will learn about the Normal Distribution. You will learn what it is and why it is important, and you willbegin to develop an intuition for the rarity of a value in a set by comparing it to the mean and standard deviation ofthe data.
If you knew that the prices of t-shirts sold in an online shopping site were normally distributed, and had a meancost of $10, with a standard deviation of $1.50, how could that information benefit you as you are looking at varioust-shirt prices on the site? How could you use what you know if you were looking to make a profit by purchasing unusually inexpensive shirts to resell at prices that are more common?
Watch This: Spread of a Normal Distribution
Guidance
A distribution is an evaluation of the way that points in a data set are clustered or spread across their range ofvalues. A normal distribution is a very specific symmetrical distribution that indicates, among other things, thatexactly [latex]\frac{1}{2}\\[/latex]of the data is below the mean, and [latex]\frac{1}{2}\\[/latex]is above, that approximately 68% of the data is within 1, approximately96% of the data is within 2, and approximately 99.7% is within 3 standard deviations of the mean.
There are a number of reasons that it is important to become familiar with the normal distribution, as you willdiscover throughout this chapter. Examples of values associated with normal distribution:
- Physical characteristics such as height, weight, arm or leg length, etc.
- The percentile rankings of standardized testing such as the ACT and SAT
- The volume of water produced by a river on a monthly or yearly basis
- The velocity of molecules in an ideal gas
Knowing that the values in a set are exactly or approximately normally distributed allows you to get a feel for howcommon a particular value might be in that set. Because the values of a normal distribution are predictably clusteredaround the mean, you can estimate in short order the rarity of a given value in the set. In our upcoming lesson onthe Empirical Rule, you will see that it is worth memorizing that normally distributed data has the characteristicsmentioned above:
- 50% of all data points are above the mean and 50% are below
- Apx 68% of all data points are within 1 standard deviation of the mean
- Apx 95% of all data points are within 2 standard deviations of the mean
- Apx 99.7% of all data points are within 3 standard deviations of the mean
In this lesson, we will be practicing a ’rough estimate’ of the probability that a value within a given range willoccur in a particular set of data, just to develop an intuition of the use of a normal distribution. In subsequentlessons, we will become more specific with our estimates. The image below will be used in greater detail in thelesson on the Empirical Rule, but you may use it as a reference for this lesson also.
Example 1
Human height is commonly considered an approximately normally distributed measure. If the mean height of a male adult in the United States is 5′ 10″, with a standard deviation of 1.5″, how common are men with heights greater than 6′ 2″?
Solution
Since each standard deviation of this normally distributed data is 1.5″, and 6′ 2″ is 4″ above the mean for the population, 6′ 2″ is nearly 3 standard deviations above the mean. That tells us that men taller than 6′ 2″ are quite rare in this population.
Example 2
If the fuel mileage of a particular model of car is normally distributed, with a mean of 26 mpg and a standarddeviation of 2 mpg, how common are cars with a fuel efficiency of 24 to 25 mpg?
Solution
We know that apx 68% of the cars in the population have an efficiency of between 24 and 28 mpg, since that wouldbe 1 SD below and 1 SD above the mean. That suggests that apx 34% have an efficiency of 24 to 26 mpg, so we cansay that it is uncommon to see a car with an efficiency between 24 and 25 mpg, but not extremely so.
Example 3
If the maximum jumping height of US high school high jumpers is normally distributed with a mean of 5′ 11.5″ anda SD of 2.2″, how unusual is it to see a high school jumper clear 6′ 3″?
Solution
If the mean is 5′ 11.5″, then 1 SD above is 6′ 1.7″ and 2 SDs is 6′ 3.9″. That means that less than 2.5% of jumpers6′3.9″, so it would be pretty uncommon to see a high-school competitor exceed 6′3″.
Intro Problem Revisited
If you knew that the prices of t-shirts sold in an online shopping site were normally distributed, and had a meancost of $10, with a standard deviation of $1.50, how could that information benefit you as you are looking at varioust-shirt styles and designs on the site? How could you use what you know if you were looking to make a profit bypurchasing unusually inexpensive shirts to resell at prices that are more common?
By knowing the mean and SD of the shirt prices, and knowing that they are normally distributed, you can estimateright away if a shirt is priced at a point significantly below the norm. For instance, with this data, we can estimatethat a shirt priced at $7.00 is less expensive than apx 97.5% of all shirts on the site, and could likely be resold at aprofit (assuming there is not something wrong the shirt that is not obvious from the listing).
Vocabulary
Distribution: an arrangement of values of a variable showing their observed or theoretical frequency of occurrence.
Range of values of a distribution: is the difference between the least and greatest values.
Normal distribution: a very specific distribution that is symmetric about its mean. Half the values of therandom variable are below the mean and half are above the mean. Approximately 68% of the data is within1 standard deviation of the mean;aproximately 96% is within 2 SDs, and 99.7% within 3 SDs.
Standard deviation:a measure of how spread out the data is from the mean. To determine if a data value is farfrom the mean, determine how many standard deviations it is from the mean. The SD is calculated as the square rootof the variance.
Guided Practice
Assume the data to be normally distributed, and describe the rarity of an event using the followingscale:
- 0% to< 1% probability = very rare
- 1% to< 5% = rare
- 5% to< 34% = uncommon
- 34% to< 50% = common
- 50% to100% = likely
Questions
- If the mean (µ) of the data is 75, and the standard deviation (σ) is 5, how common is a value between 70 and75?
- If the µ is .02 and the σ is .005, how common is a value between .005 and .01?
- If the µ is 1280 and the σ is 70, how common is a value between 1210 and 1350?
- If the mean defect rate at a cellphone production plant is .1%, with a standard deviation of .03%, would itseem reasonable for a quality assurance manager to be concerned about 3 defective phones in a single 1000unit run?
Solutions
- A value of 70 is only 1 standard deviation below the mean, so a value between 70 and 75 would be expectedapproximately 34% of the time, so it would be common.
- A value of .01 is 2 SDs below the mean, and .005 is 3 SDs below, so we would expect there to be about a2.5% probability of a value occurring in that range. A value between 0.005 and 0.01 would be rare.
- 1210 is 1 SD below the mean, and 1350 is 1 SD above the mean, so we would expect approximately 68% ofthe data to be in that range, meaning that it is likely that a value in that range would occur.
- .1% translates into 1 per thousand, with a standard deviation of 3 per ten thousand. That means that 3 defects inthe same thousand is nearly 7 SDs above the mean, well into the very rare category. While it is not impossiblefor random chance to result in such a value, it would certainly be prudent for the manager to investigate.
Practice Questions
Assume all sets/populations to be approximately normally distributed, and describe the rarity of an event usingthe following scale:
- 0%to< 1% probability = very rare
- 1% to < 5% = rare
- 5% to < 34% = uncommon
- 34% to <50% = common
- 50% to 100% = likely.
You may reference the image below:
Attributions
This chapter contains material taken from Math in Society (on OpenTextBookStore) by David Lippman, and is used under a CC Attribution-Share Alike 3.0 United States(CC BY-SA 3.0 US) license.
This chapter contains material taken from of Math for the Liberal Arts (on Lumen Learning) by Lumen Learning, and is used under a CC BY: Attribution license.
