5.3: Connecting the Standard Deviation and Normal Distribution
Learning Objectives
 Represent the standard deviation of a normal distribution on the bell curve.
 Use the percentages associated with normal distribution to solve problems.
Introduction
In the problem presented in lesson one, regarding your test mark, your teacher told you that the class marks were normally distributed. In the previous lesson you calculated the standard deviation of the marks by using the TI83 calculator. Later in this lesson, you will be able to represent the value of the standard deviation as it relates to a normal distribution curve.
You have already learned that 68% of the data lies within one standard deviation of the mean, 95% of the data lies within two standard deviations of the mean and 99.7% of the data lies within three standard deviations of the mean. To accommodate these percentages, there are defined values in each of the regions to the left and to the right of the mean.
These percentages are used to answer real world problems when both the mean and the standard deviation of a data set are known.
Example 1:
The lifetimes of a certain type of calculator battery are normally distributed. The mean life is 400 hours, and the standard deviation is 50 hours. For a group of 5000 batteries, how many are expected to last
a) between 350 hours and 450 hours?
b) more than 300 hours?
c) less than 300 hours?
Solution:
a) 68% of the batteries lasted between 350 hours and 450 hours. This means that \begin{align*}(5000 \times .68 = 3400)\end{align*}
b) \begin{align*}95\% + 2.35\% = 97.35\%\end{align*}
c) Only 2.35% of the batteries are expected to last less than 300 hours. This means that \begin{align*}(5000 \times .0235=117.5 \approx 118)\end{align*}
Example 2:
A bag of chips has a mean mass of 70 g with a standard deviation of 3 g. Assuming normal distribution; create a normal curve, including all necessary values.
a) If 1250 bags are processed each day, how many bags will have a mass between 67g and 73g?
b) What percentage of chips will have a mass greater than 64g?
Solution:
a) Between 67g and 73g, lies 68% of the data. If 1250 bags of chips are processed, 850 bags will have a mass between 67 and 73 grams.
b) 97.35% of the bags of chips will have a mass greater than 64 grams.
Now you can represent the data that your teacher gave to you for your recent Math test on a normal distribution curve. The mean mark was 61 and the standard deviation was 15.6.
From the normal distribution curve, you can say that your mark of 71 is within one standard deviation of the mean. You can also say that your mark is within 68% of the data. You did very well on your test.
Lesson Summary
In this chapter you have learned what is meant by a set of data being normally distributed and the significance of standard deviation. You are now able to represent data on the bellcurve and to interpret a given normal distribution curve. In addition, you can calculate the standard deviation of a given data set both manually and by using technology. All of this knowledge can be applied to real world problems which you are now able to answer.
Points to Consider
 Is the normal distribution curve the only way to represent data?
 The normal distribution curve shows the spread of the data but does not show the actual data values. Do other representations of data show the actual data values?
Review Questions
 Without using technology, calculate the variance and the standard deviation of each of the following sets of numbers.
 2, 4, 6, 8, 10, 12, 14, 16, 18, 20
 18, 23, 23, 25, 29, 33, 35, 35
 123, 134, 134, 139, 145, 147, 151, 155, 157
 58, 58, 65, 66, 69, 70, 70, 76, 79, 80, 83
 Ninetyfive percent of all cultivated strawberry plants grow to a mean height of 11.4 cm with a standard deviation of 0.25 cm.
 If the growth of the strawberry plant is a normal distribution, draw a normal curve showing all the values.
 If 225 plants in the greenhouse have a height between 11.15 cm and 11.65 cm, how many plants were in the greenhouse?
 How many plants in the greenhouse would we expect to be shorter than 10.9 cm?
 The coach of the high school basketball team asked the players to submit their heights. The following results were recorded. \begin{align*}&175 \ cm && 179 \ cm && 179 \ cm && 181 \ cm && 183 \ cm\\
&183 \ cm && 184 \ cm && 184 \ cm && 185 \ cm && 187 \ cm\end{align*}
175 cm183 cm179 cm184 cm179 cm184 cm181 cm185 cm183 cm187 cm Without using technology, calculate the standard deviation of this set of data.  A survey was conducted at a local high school to determine the number of hours that a student studied for the final Math 10 exam. To achieve a normal distribution, 325 students were surveyed. The results showed that the mean number of hours spent studying was 4.6 hours with a standard deviation of 1.2 hours.
 Draw a normal curve showing all the values.
 How many students studied between 2.2 hours and 7 hours?
 What percentage of the students studied for more than 5.8 hours?
 Harry noticed that he scored a mark of 60 on the Math 10 exam but had studied for \begin{align*}\frac{1}{2}\end{align*}
12 hour. Is Harry a typical student? Explain.
 A group of grade 10 students at one high school were asked to record the number of hours they watched television per week, the results are recorded in the table shown below \begin{align*}& 2.5 && 3 && 4.5 && 4.5 && 5 && 5 && 5.5 && 6 && 6 && 7\\
&8 && 9 && 9.5 && 10 && 10.5 && 11 && 13 && 16 && 26 && 28\end{align*}
2.58394.59.54.510510.55115.513616626728 Using Technology (TI83), calculate the variance and the standard deviation of this data.  The average life expectancy for a dog is 10 years 2 months with a standard deviation of 9 months.
 If a dog’s life expectancy is a normal distribution, draw a normal curve showing all values.
 What would be the lifespan of almost all dogs? (99.7%)
 In a sample of 825 dogs, how many dogs would have life expectancy between 9 years 5 months and 10 years 11 months?
 How many dogs, from the sample, would we expect to live beyond 10 years 11 months?
 Ninetyfive percent of all Marigold flowers have a height between 10.9 cm and 119.0 cm and their height is normally distributed.
 What is the mean height of the Marigolds?
 What is the standard deviation of the height of the Marigolds?
 Draw a normal curve showing all values for the heights of the Marigolds.
 If 208 flowers have a height between 11.15 cm and 11.65 cm, how many flowers were in our sample?
 How many flowers in our sample would we expect to be shorter than 10.9 cm?
 A group of physically active women were asked to record the number of hours they spent at the gym each week. The results are shown below. \begin{align*}&8 && 8 && 9 && 9 && 9 && 9.5 && 9.5 && 9.5 && 9.5 && 9.5\\
&9.5 && 9.5 && 9.5 && 9.5 && 9.5 && 10 && 10 && 10 && 11 && 11\end{align*}
89.589.599.599.599.59.5109.5109.5109.5119.511 Calculate the standard deviation.  A normal distribution curve shows a mean \begin{align*}(\bar x)\end{align*} and a standard deviation \begin{align*}(\sigma)\end{align*}. Approximately what percentage of the data would lie in the intervals with the limits shown?
 \begin{align*}\bar x  2 \sigma, \bar x + 2 \sigma\end{align*}
 \begin{align*}\bar x, \bar x + 2 \sigma\end{align*}
 \begin{align*}\bar x  \sigma, \bar x + \sigma\end{align*}
 \begin{align*}\bar x  \sigma, \bar x\end{align*}
 \begin{align*}\bar x  \sigma, \bar x + 2 \sigma\end{align*}
 Use the 689599.7 rule on a normal distribution of data with a mean of 185 and a standard deviation of 10, to answer the following questions. What percentage of the data would measure
 between 175 and 195?
 between 195 and 205?
 between 155 and 215?
 between 165 and 185?
 between 185 and 215?
Review Answers

 \begin{align*}\sigma^2 = 33 \ \ \sigma = 5.74\end{align*}
 \begin{align*}\sigma^2 = 35.24 \ \ \sigma = 5.94\end{align*}
 \begin{align*}\sigma^2 = 111.28 \ \ \sigma = 10.55\end{align*}
 \begin{align*}\sigma^2 = 64.96 \ \ \sigma = 8.06\end{align*}
(b) 68% of the plants have a height between 11.15 cm and 11.65 cm. \begin{align*}0.68 (x) & = 225\\ x & = \frac{225}{0.68}\\ x & = 331\end{align*}
Therefore there were 331 strawberry plants in the greenhouse.
(c) \begin{align*}& \quad \ 99.7 \% \quad  \quad 95 \% = 4.7 \%\\ & \qquad \Bigg\uparrow \qquad \qquad \quad \Bigg\uparrow \\ & \ \text{All plants} \qquad \text{plants with}\\ & \ \text{within} \ 3 \sigma \qquad \text{heights greater}\\ & \ \text{from mean.} \quad \ \text{than} \ 10.9 \ \text{cm}\\ & \frac{4.7 \%}{2} = 2.35 \%\\ & 331 \times 0.0235 = 8 \ \text{plants}\end{align*}
Therefore, eight plants in the greenhouse would be shorter than 10.9 cm.
\begin{align*}x\end{align*}  \begin{align*}(x  \bar x)\end{align*}  \begin{align*}(x  \bar x)^2\end{align*} 

175  7  49 
179  3  9 
179  3  9 
181  1  1 
183  1  1 
183  1  1 
184  2  4 
184  2  4 
185  3  9 
187  5  25 
Sum = 1820  112 
\begin{align*}\bar x & = \frac{1820}{10} && \sigma = \sqrt{ \frac{\sum (x  \bar x)^2}{n}} && \sigma = \sqrt{ \frac{112}{10}} && \sigma = \sqrt{11.2}\\ \bar x & = 182 &&&& \sigma = 3.35\end{align*}
(a)
(b) \begin{align*}95 \% \ \text{of students} = 0.95 \times 325 \ \text{students} = 308 \ \text{students}\end{align*}
Therefore 308 students studied between 2.2 and 7 hours.
(c) \begin{align*}\frac{1}{2}(99.7 \%  68 \%) & = \frac{1}{2}(31.7 \%)\\ & =15.85 \%\end{align*}
15.85 % of the students studied longer than 5.8 hours.
(d) Harry is not a typical student. The mean is 4.6 hours; therefore the majority of students studied more than 4 hours more than Harry did for the exam. Harry is lucky to have received a 60% on the exam.
The standard deviation of the data is approximately 6.72 and the variance is approximately 45.18. This large variation in the data is described by the larger standard deviation.
(a)
(b) Almost all dogs have a life span of 7 years 11 months to 12 years 5 months.
(c) \begin{align*}34 \% + 34 \% & = 68 \%\\ (0.68 \times 825 & = 561)\end{align*}
In a sample of 825 dogs, 561 would have a life expectancy between 9 years 5 months to 10 years 11 months.
(d) \begin{align*}13.5 \% + 2.35 \% & = 15.85 \%\\ 0.1585 \times 825 & = 130.76\end{align*}
In a sample of 825 dogs, 130 would have a life expectancy of more than 10 years 11 months.
 (a) (11.4 cm) (b) (0.25) (c) Normal Distribution Curve (d) There are 306 flowers in the sample. (e) Seven flowers would be shorter than 10.9 cm.
Data \begin{align*}x\end{align*}  Mean\begin{align*}(\bar x)\end{align*} 
(Data – Mean) \begin{align*}(x  \bar x)\end{align*} 
(Data  Mean)\begin{align*}^2\end{align*} \begin{align*}(x  \bar x)^2\end{align*} 

8  9.5  1.5  2.25 
8  9.5  1.5  2.25 
9  9.5  0.5  0.25 
9  9.5  0.5  0.25 
9  9.5  0.5  0.25 
9.5  9.5  0  0 
9.5  9.5  0  0 
9.5  9.5  0  0 
9.5  9.5  0  0 
9.5  9.5  0  0 
9.5  9.5  0  0 
9.5  9.5  0  0 
9.5  9.5  0  0 
9.5  9.5  0  0 
9.5  9.5  0  0 
10  9.5  0.5  0.25 
10  9.5  0.5  0.25 
10  9.5  0.5  0.25 
11  9.5  1.5  2.25 
11  9.5  1.5  2.25 
190  10.5 
\begin{align*}\overline{x} & = 9.5\\ \sigma & = 0.72\end{align*}
 (95%)
 (47.5%)
 (68%)
 (34%)
 (81.5%)
 68%
 13.5%
 99.7%
 47.5%
 48.85%
Answer Key for Review Questions (even numbers)
2.
(b) 68% of the plants have a height between 11.15 cm and 11.65 cm. \begin{align*}0.68 (x) & = 225\\ x & = \frac{225}{0.68}\\ x & = 331\end{align*}
Therefore there were 331 strawberry plants in the greenhouse.
(c) \begin{align*}& \quad \ 99.7 \% \quad  \quad 95 \% = 4.7 \%\\ & \qquad \Bigg\uparrow \qquad \qquad \quad \Bigg\uparrow \\ & \ \text{All plants} \qquad \text{plants with}\\ & \ \text{within} \ 3 \sigma \qquad \text{heights greater}\\ & \ \text{from mean.} \quad \ \text{than} \ 10.9 \ \text{cm}\\ & \frac{4.7 \%}{2} = 2.35 \%\\ & 331 \times 0.0235 = 8 \ \text{plants}\end{align*}
Therefore, eight plants in the greenhouse would be shorter than 10.9 cm.
4. (a)
(b) \begin{align*}95 \% \ \text{of students} = 0.95 \times 325 \ \text{students} = 308 \ \text{students}\end{align*}
Therefore 308 students studied between 2.2 and 7 hours.
(c) \begin{align*}\frac{1}{2}(99.7 \%  68 \%) & = \frac{1}{2}(31.7 \%)\\ & =15.85 \%\end{align*}
15.85 % of the students studied longer than 5.8 hours.
(d) Harry is not a typical student. The mean is 4.6 hours; therefore the majority of students studied more than 4 hours more than Harry did for the exam. Harry is lucky to have received a 60% on the exam.
6. (a)
(b) Almost all dogs have a life span of 7 years 11 months to 12 years 5 months.
(c) \begin{align*}34 \% + 34 \% & = 68 \%\\ (0.68 \times 825 & = 561)\end{align*}
In a sample of 825 dogs, 561 would have a life expectancy between 9 years 5 months to 10 years 11 months.
(d) \begin{align*}13.5 \% + 2.35 \% & = 15.85 \%\\ 0.1585 \times 825 & = 130.76\end{align*}
In a sample of 825 dogs, 130 would have a life expectancy of more than 10 years 11 months.
8.
Data \begin{align*}x\end{align*}  Mean\begin{align*}(\bar x)\end{align*}  (Data – Mean) \begin{align*}(x  \bar x)\end{align*}  (Data  Mean)\begin{align*}^2\end{align*} \begin{align*}(x  \bar x)^2\end{align*} 

8  9.5  1.5  2.25 
8  9.5  1.5  2.25 
9  9.5  0.5  0.25 
9  9.5  0.5  0.25 
9  9.5  0.5  0.25 
9.5  9.5  0  0 
9.5  9.5  0  0 
9.5  9.5  0  0 
9.5  9.5  0  0 
9.5  9.5  0  0 
9.5  9.5  0  0 
9.5  9.5  0  0 
9.5  9.5  0  0 
9.5  9.5  0  0 
9.5  9.5  0  0 
10  9.5  0.5  0.25 
10  9.5  0.5  0.25 
10  9.5  0.5  0.25 
11  9.5  1.5  2.25 
11  9.5  1.5  2.25 
190  10.5 
\begin{align*}\overline{x} & = 9.5\\ \sigma & = 0.72\end{align*}
10. (a) 68%
(b) 13.5%
(c) 99.7%
(b) 47.5%
(d) 48.85%
Vocabulary
 Normal Distribution
 A symmetric bellshaped curve with tails that extend infinitely in both directions from the mean of a data set.
 Standard Deviation
 A measure of spread of the data equal to the square root of the sum of the squared variances divided by the number of data.
 Variance
 A measure of spread of the data equal to the mean of the squared variation of each data value from the mean.
 689599.7 Rule
 The percentages that apply to how the standard deviation of the data spreads out from the mean of a set of data.
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