Normal Distributions

Key Concepts

• Understand how to find population intervals
• Understand how to use the empirical rule
• Understand how to compare value using z-scores
• Understand how to use a a-score to compute percentage 

Explore and Reason 

The owner of an apple orchard and the owner of an orange grove create histograms to display fruit production data. 

1. Describe the shape of each distribution. Discuss how the distributions are alike and how they are different. 
2. Use structure: 

Explain how you could estimate the mean from the graphs, the standard deviation of the mean on each graph. 

Solution: 

1. Both the histograms are symmetric.  
2. I am going to follow the following strategies: 

Mean = ∑xifi / n where, 

xi= mid value of the class intervals 

fi= Frequency 

n = total frequency 

Median = l +

n/2 −c f / f*x h where, 

L = lower boundary of the median class  

n = total frequency  

cf = cumulative frequency of the median class 

f = frequency of the median class 

h = size of the class   

Concept: The Empirical rule 

The normal distribution has a special property called the empirical rule. The approximate percentage of data values falling in any interval of the range of the data can be determined using just the mean and the standard deviation. 

How far from the mean does a value in a normal distribution fall? 

• About 68% of all values fall within 1 standard deviations 
• About 95% of all values fall within 2 standard deviations 
• About 99.7% of all values fall within 3 standard deviations 

The rule applies only to normal distributions. 

• For a population, the mean is denoted by μ and the standard deviation by σ 

Example 1: Find population intervals 

An example of an early application of statistics was in the year 1817. A study of chest  

circumference among a group of Scottish men exhibited an approximately normal distribution. 

Their chest circumferences ranged from 33 to 48 in., with a mean chest measurement of 40 in., and 

a standard deviation of 2 in. Use the empirical rule to help you understand the distribution of 

chest circumferences in the study. 

1. What range of chest measurements contains the 68% which fall closest to the mean? 

To draw the graph, first draw the horizontal axis. Draw the ‘bell curve’ of the normal distribution and add the labels to the axis with the mean at the centre. Count by the standard deviation to each end of the curve. 

You would expect approximately 68% of the men in the population had chest measurements 

 between 38 and 42 in. 

2. What would you expect the chest measurements to be for the 2.5% of the men with the smallest chest measurements in the population?  

You would expect 2.5% of the men in the population to have chests measuring less than  

36 in. 

Try it  

1. What would you expect to be the smallest and largest chest measurements of the ‘middle’ 95% of the men? 
2. What would you expect to be the measurements of the 16% of the men with the largest chests in the population? 

Solution: 

1. From the question, the mean is 40 in and the standard deviation is 2. 

95% of all values fall within 2 standard deviations, 

μ – 2σ = 40 – 2(2)  

=40 – 4   

=36 

μ + 2σ = 40 + 2(2)  

=40 + 4 

=44 

95% of all values fall within the range 36 and 44. 

2. What would you expect to be the measurements of the 16% of the men with the largest chests in the population? 

Solution: 

Total percentage of normal distribution is 100%. 

68% of data fall within 1 standard deviation.  

So, remaining 32% cover remaining data.  

16% of the men with the largest chests in the population fall above 42 in. 

Example 2: Use the Empirical Rule 

Recent SAT scores for college-bound seniors are normally distributed with a mean score of 508  

and a standard deviation of 121. How does the empirical rule help you understand the performances of individual students? 

1. What proportion of students received scores between 387 and 629? 

Sketch the graph to see where the numbers 387 and 629 fall in terms of the distribution.  

According to the empirical rule, 68% of scores are within 1 standard deviation of the mean. That the mean is that approximately 68% of students received SAT math scores 387 and 629. 

2. What proportion of students received scores greater than 387? 

So, 68% + 16% = 84% of students received scores greater than 387. 

Try it: 

2. Find the proportion of students who earned SAT Math  

scores in the following ranges: 

1. Between 266 and 750 
2. Between 266 and 629 

Solution: 

1. According to empirical rule, 95% of scores are within 2 standard deviations. So, about 95 % of SAT math scores are within the range 266 and 750. 
2. According to the empirical rule, Data values from 387 to 629 are within 1 standard deviation. Data values from 266 to 387 are within 2^nd standard deviation. 13% of data is in the range 266 to 387. 

Then,  

From 266 to 629 = 13% + 68% 

                            = 81% 

Example 3: Compare Values using Z-scores  

Ella and Alicia comparing scores on their college entrance exams. Ella’s SAT score is 1380. 

Alicia’s ACT score is 32. The mean SAT score is 1000 with a standard deviation of 200. 

The mean ACT score is 21 with a standard deviation of 5. Who has the better score? 

You cannot compare their scores directly, but you can compare their relative’s position within each distribution. 

To analyze where a data value falls in the distribution, find how many standard deviations 

It is above or below the mean. Most data values are not an exact integer number of standard deviations from the mean, so we need to find a way to describe fractional numbers of standard deviations from the mean. 

The z-score counts how many standard deviations a data value is above or below the mean, because it divides the difference from the mean by the distance of a standard deviation. These distributions are normal, but you can use z-scores with any distributions. 

Alicia’s score is better than Ella’s.  

Try it 

3. How does an SAT score of 1120 compare to an ACT score of 23? 

Solution:  

Mean and standard deviation of SAT scores are 1000 and 32, respectively. 

So, 

z = data value −mean / standard deviation

= 1120 −1000 / 32

= 120 / 32

= 3.75   

Mean and standard deviation of ACT scores are 21 and 5, respectively. 

So, 

z =data value −mean / standard deviation

= 23 − 21 / 5

=2 / 5

= 0.4   

Therefore, z-score of SAT is better than ACT score 

Example 4: Use a z-Score to Compute Percentage 

The weight of captive adult female lowland gorillas is normally distributed with a mean of μ = 82.30 kg and standard deviation of σ = 15.33 kg. Calculate a z-score for Sasha. How  can you use this score to find the percentage of captive adult female lowland gorillas whose weights are less than or equal to Sasha’s? 

Sasha’s weight of 61.8 kg gives her a z-score of

61.8 −82.3 / 15.33 ≈ – 1.34 

When you calculate the z-score for a data value, you are finding its corresponding value on the standard normal distribution. The standard normal distribution is the normal distribution with mean μ = 0 and standard deviation σ = 1. 

The total area under the curve of the standard normal distribution is equal to 1, and represents 100% of the data values. 

The percentage of values less than or equal to a particular data value is equal to the percentage of the total area under the distribution curve for that population to the left of that data value.  

This percentage is called the percentile of the data value.  

The percentage of area to the left of a data value is equal to the area to the left of the value’s z-score under the standard normal distribution. 

To calculate the area under the curve of the standard normal distribution to the left of z = -1.34, use a spreadsheet program or a calculator. 

The percentile of Sasha’s weight or of a z-score of -1.34 is approximately 9%.  

This means that about 9% of all captive adult female lowland gorillas have a weight less than or equal to Sasha’s weight. 

Concept Summary 

Using normal distributions in the real world 

Empirical Rule

Approximately 68% of data values in a normal distribution fall within 1 standard deviation of the mean. 

Approximately 95% of data values in a normal distribution fall within 2 standard deviations of the mean. 

Approximately 99.7% of data values in a normal distribution fall within 3 standard deviations of the mean. 

z-SCORES 

z = data value −mean / standard deviation 

z tells how many standard deviations a data value is above or below the mean. 

Standard Normal Distribution

Mean: 0  

Standard Deviation: 1 

Along with z-scores, the standard normal distribution allows you to compare values across different population distributions.  

Normal distributions are predictable, allowing you to calculate percentiles using tables, calculators, or spreadsheets. 

Let’s check your knowledge

The life span of a certain brand of car tire is approximately normally distributed. The car tires have a mean life span of 50,000 miles and a standard deviation of 7500 miles. 

1. What range of car tire life span contains the 95% closet to the mean? 
2. What would be the lifespan be for the 2.5% of the tires with the greatest life span in the population? 

The price of a certain brand of printers is normally distributed with the mean of $215 and standard deviation $32. 

1. What proportion of printer’s cost between $110 and $320?  
2. What proportion of printer’s cost less than $145?  
3. What proportion of printer’s cost more than $250?  

Find the z-score with the given data: 

1. μ = 0 , σ = 2, x =3  
2. μ = 1 , σ = 0.15, x =0.70  
3. μ = 100 , σ = 15, x =70  

Answers 

1. 

Solution: 

1. 95% of the data values are within 2 standard deviations.  

μ – 2σ = 50000 – 2(7500) 

=50000 – 15000  

=35000 

μ + 2σ = 50000 + 2(7500) 

=50000 + 15000 

=65000 

95% of the data values are within the range 35000 to 65000. 

2. What would be the lifespan for the 2.5% of the tires with the greatest life span in the population? 

Solution: 

μ + 2σ = 50000 + 2(7500) 

              =50000 + 15000 

              =65000 

65000 is the life span for 2.5% of the tires with the greatest life span in the population. 

2. 

Solution: 

a.   

Range is $110 and $320   

Let ‘K’ be the nth standard deviation   

215 – k (32) = 110  

          – 32k = – 105 

                  k = [105/32] 

                   k ≈ 3 

The proportion of printers cost between $110 and $320 is 99.7% 

2. What proportion of printers cost less than $145?  

Solution: 

2.35 % of the printer’s cost is less than $145  

3. What proportion of printer’s cost more than $250? 

Solution: 

16% of the printer’s cost is more than $250. 

3. 

Solution: 

a.  

μ = 0 , σ = 2 , x =3   

z = data value −mean / standard deviation 

= 3 −0 / 2

= 3 / 2 = 1.5 

b. 

μ = 1 , σ = 0.15 , x =0.70  

z = data value −mean / standard deviation  

= 0.70 − 1/0.1 = −0.30/0.15

= – 2 

c.   

μ = 100 , σ = 15 , x =70  

z = data value −mean / standard deviation 

=70 −100/15

=30/15  = 2 

Exercise

• The graph of normally distributed data is shown. What are the mean and standard deviation of the data? Explain how you know.

• Find the percentage of all the values in a normal distribution for each z-score.
• z ≤ 2.15
• z ≤ 1.25
• z ≥ -1.39
• Explain how to use the empirical rule to find the percentage of the population that falls in a given interval of values.

Concept Summary