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# 11.9: The Empirical Rule

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### Objective

Here you will learn how to use the Empirical Rule to estimate the probability of an event.

If the price per pound of USDA Choice Beef is normally distributed with a mean of $4.85/lb and a standard deviation of$0.35/lb, what is the estimated probability that a randomly chosen sample (from a randomly chosen market) will be between $5.20 and$5.55 per pound?

### Watch This: Empirical Rule

A Vimeo element has been excluded from this version of the text. You can view it online here: http://pb.libretexts.org/mala/?p=156

## Guidance

This reading on the Empirical Rule is an extension of the previous reading “Understanding the Normal Distribution.” In the prior reading, the goal was to develop an intuition of the interaction between decreased probability and increased distance from the mean. In this reading, we will practice applying the Empirical Rule to estimate the specific probability of occurrence of a sample based on the range of the sample, measured in standard deviations.

The graphic below is a representation of the Empirical Rule:

The graphic is a rather concise summary of the vital statistics of a Normal Distribution. Note how the graph resembles a bell? Now you know why the normal distribution is also called a “ bell curve.”

• 50% of the data is above, and 50% below, the mean of the data
• Approximately 68% of the data occurs within 1 SD of the mean
• Approximately 95% occurs within 2 SD’s of the mean
• Approximately 99.7% of the data occurs within 3 SDs of the mean

It is due to the probabilities associated with 1, 2, and 3 SDs that the Empirical Rule is also known as the 68−95−99.7 rule.

### Example 1

If the diameter of a basketball is normally distributed, with a mean (µ) of 9″, and a standard deviation (σ) of 0.5″, what is the probability that a randomly chosen basketball will have a diameter between 9.5″ and 10.5″?

#### Solution

Since the σ = 0.5″ and the µ = 9″, we are evaluating the probability that a randomly chosen ball will have a diameter between 1 and 3 standard deviations above the mean. The graphic below shows the portion of the normal distribution included between 1 and 3 SDs:

The percentage of the data spanning the 2nd and 3rd SDs is 13.5% + 2.35% = 15.85%

The probability that a randomly chosen basketball will have a diameter between 9.5 and 10.5 inches is 15.85%.

### Example 2

If the depth of the snow in my yard is normally distributed, with µ = 2.5″ and σ = .25″, what is the probability that a randomly chosen location will have a snow depth between 2.25 and 2.75 inches?

#### Solution

2.25 inches is µ − 1σ, and 2.75 inches is µ + 1σ, so the area encompassed approximately represents 34% + 34% = 68%.

The probability that a randomly chosen location will have a depth between 2.25 and 2.75 inches is 68%.

### Example 3

If the height of women in the United States is normally distributed with µ = 5′ 8″ and σ = 1.5″, what is the probability that a randomly chosen woman in the United States is shorter than 5′ 5″?

#### Solution

This one is slightly different, since we aren’t looking for the probability of a limited range of values. We want to evaluate the probability of a value occurring anywhere below 5′ 5″. Since the domain of a normal distribution is infinite, we can’t actually state the probability of the portion of the distribution on “that end” because it has no “end”! What we need to do is add up the probabilities that we do know and subtract them from 100% to get the remainder.

Here is that normal distribution graphic again, with the height data inserted:

Recall that a normal distribution always has 50% of the data on each side of the mean. That indicates that 50% of US females are taller than 5′ 8″, and gives us a solid starting point to calculate from. There is another 34% between 5′ 6.5″ and 5′ 8″ and a final 13.5% between 5′ 5″ and 5′ 6.5″. Ultimately that totals: 50% + 34% + 13.5% = 97.5%. Since 97.5% of US females are 5′ 5″ or taller, that leaves 2.5% that are less than 5′ 5″ tall.

## Intro Problem Revisited

If the price per pound of USDA Choice Beef is normally distributed with a mean of $4.85/lb and a standard deviation of$0.35/lb, what is the estimated probability that a randomly chosen sample (from a randomly chosen market) will  be between $5.20 and$5.55 per pound?

$5.20 is µ + 1σ, and$5.55 is µ + 2σ, so the probability of a value occurring in that range is approximately 13.5%.

### Vocabulary

Normal distribution: a common, but specific, distribution of data with a set of characteristics detailed in the lesson above.

Empirical Rule: a name for the way in which the normal distribution divides data by standard deviations: 68% within 1 SD, 95% within 2 SDs and 99.7 within 3 SDs of the mean

68-95-99.7 rule: another name for the Empirical Rule

Bell curve: the shape of a normal distribution

### Guided Practice

1. A normally distributed data set has µ = 10 and σ = 2.5, what is the probability of randomly selecting a value greater than 17.5 from the set?
2. A normally distributed data set has µ = .05 and σ = .01, what is the probability of randomly choosing a value between .05 and .07 from the set?
3. A normally distributed data set has µ = 514 and an unknown standard deviation, what is the probability that a randomly selected value will be less than 514?

#### Solutions

1. If µ = 10 and σ = 2.5, then 17.5 = µ + 3σ. Since we are looking for all data above that point, we need to subtract the probability that a value will occur below that value from 100%: The probability that a value will be less than 10 is 50%, since 10 is the mean. There is another 34% between 10 and 12.5, another 13.5% between 12.5 and 15, and a final 2.35% between 15 and 17.5. 100% −50% −34% −13.5% −2.35% = 0.15% probability of a value greater than 17.5
2. 0.05 is the mean, and 0.07 is 2 standard deviations above the mean, so the probability of a value in that range is 34% + 13.5% = 47.5%
3. 514 is the mean, so the probability of a value less than that is 50%.

## Practice Questions

Assume all distributions to be normal or approximately normal, and calculate percentages using the 68−95−99.7 rule.

1. Given mean 63 and standard deviation of 168, find the approximate percentage of the distribution that lies between −105 and 567.
2. Approximately what percent of a normal distribution is between 2 standard deviations and 3 standard deviations from the mean?
3. Given standard deviation of 74 and mean of 124, approximately what percentage of the values are greater than 198?
4. Given σ = 39 and µ = 101, approximately what percentage of the values are less than 23?
5. Given mean 92 and standard deviation 189, find the approximate percentage of the distribution that lies between −286 and 470.
6. Approximately what percent of a normal distribution lies between µ + 1σ and µ + 2σ?
7. Given standard deviation of 113 and mean 81, approximately what percentage of the values are less than −145?
8. Given mean 23 and standard deviation 157, find the approximate percentage of the distribution that lies between 23 and 337.
9. Given σ = 3 and µ = 84, approximately what percentage of the values are greater than 90?
10. Approximately what percent of a normal distribution is between µ and µ+1σ?
11. Given mean 118 and standard deviation 145, find the approximate percentage of the distribution that lies between −27 and 118.
12. Given standard deviation of 81 and mean 67, approximately what percentage of values are greater than 310?
13. Approximately what percent of a normal distribution is less than 2 standard deviations from the mean?
14. Given µ + 1σ = 247 and µ + 2σ = 428, find the approximate percentage of the distribution that lies between 66 and 428.
15. Given µ − 1σ = −131 and µ + 1σ = 233, approximately what percentage of the values are greater than −495?