Normal Distribution
Consider the following: Four coins are flipped and the number of heads is recorded.
1. List all of the possible outcomes. Are all outcomes equally likely? Explain.
There are 16 possible outcomes.
If order is considered, then no outcome is the same. Thus,
each outcome has the same likelihood of occurring.
However, considering the number of heads possible for
each outcome, the outcomes have an unequal likelihood of
occurring.
2. What is the theoretical probability of getting 4 heads? 3 heads? 2 heads? 1 head? 0 heads?
1
4
6
4
1
4 heads= , 3 heads= , 2 heads= , 1 heads= , 0 heads=
16
16
16
16
16
Calculate the percent likelihood.
4 heads=6.25%, 3 heads=25%, 2 heads=37.5%, 1 heads=25%, 0 heads=6.25%
3. What observations can be made about the theoretical probabilities?
The data is symmetric; the fractions add up to 1; the percentages add up to 100%
4. On the axis below, complete the histogram of the probability of each number of heads.
a. Draw a point at the midpoint of the top of each bar.
Connect the points in a smooth curve.
b. What do you notice about the graph’s shape?
Symmetric, bell shaped
c. What do you observe about the graph’s mean,
median, and mode?
They are all the same (equal to 2)
Characteristics of a normal curve:
 Symmetric about the mean and bell-shaped
 Mean, median, and mode are equal
 Never crosses the x-axis
 Total area under the curve is 1 (sum of the probabilities)
Normal Distribution
Normally distributed data sets have a normal curve graph.
 Examples of normally distributed data: heights, tread life of tires, SAT scores
Normal curves follow the 68-95-99.7% Rule (aka Empirical Rule or 3 Sigma Rule)
Consider the following: Pulse rates are normally distributed with a mean of 72 and a standard
deviation of 12. Label the normal model below.
26
-3σ
48
-2σ
60
-1σ
72
µ
84
1σ
96
2σ
108
3σ
The 68-95-99.7% Rule tells us that
 68% of all people will have pulse rates within 1 standard deviation of the mean.
That means that 68% of people have pulse rates between 60 and 84
 95% of all people will have pulse rates within 2 standard deviations of the mean.
That means that 95 % of people have pulse rates between 48 and 96
 99.7% of all people will have pulse rates within 3 standard deviations of the mean.
That means that 99.7 % of people have pulse rates between 36 and 108
Values outside of 2σ or the 95% range are considered unusual.
5. Given a normally distributed data set of 500 observations measuring tree heights in a forest,
what is the approximate number of observations that fall within two standard deviations from
the mean?
Since 95% of the data falls within 2 standard deviations of the mean, then:
500 * 95% = 500 (0.95) = 475 Trees
6. A normally distributed data set containing the number of ball bearings produced during a
specified interval of time has a mean of 150 and a standard deviation of 10. What percentage
of the observed values fall between 140 and 160?
Since 140 is 1 standard deviation below the mean and 160 is 1 standard deviation above the
mean. Therefore the percentage of data that falls within 1 standard deviation is 68%.
Normal Distribution
Finding the area under a normal curve means to find the probability of a value being greater than
the rest of the data.
Areas can be found under a normal
curve by using the 68-95-99.7 rule if the
areas are bounded at places where an
exact deviation occurs.
Areas that are not bounded at ±1, ±2, or
±3 standard deviations can be found
using a z-table.
Consider the following: A corn chip factory packs chips in bags with normally distributed
weights with a mean of 12.4 oz. and a standard deviation of 0.14 oz.
7. On the graph to the right, label the mean and three
standard deviations above and below the mean.
8. Shade the region that indicates the percentage of bags
that contain less than 12.64 oz.
9. Calculate the z-score of 12.64.
12.64  12.4
z
 1.714
0.14
10. Use the z-table (Standard Normal Probabilities Table) to find the area associated with the
z-score and interpret your result.
a. The probability that a data value will fall below the data value associated with a z-score
of 1.174 is 0.9591 (95.91%).
b. 12.64 falls in the below percentile. This means that 95 percent of the data in the
distribution fall below the value associated with a z-score of 1.714
c. The probability that a value from the data set will fall above this value is
1 – 0.9591 = 0.0409 (or 4.09% of the data falls above this value)
Normal Distribution
11. On the graph at the right, label and shade the region
that represents the likelihood a bag will contain between
12.1 and 12.76 oz.
12. Calculate the z-score for each and find the probability.
12.1  12.4
z
 2.14  0.0162
0.14
z
12.76  12.4
 0.9949
0.14
13. How can we use these values to determine the probability of choosing a bag between 12.1
and 12.76 oz.?
Find the probability for each z-score. Then subtract the higher z-score’s probability from the
lower z-score’s probability.
0.9949 – 0.0162 = 0.9787 (or 97.87%)
Consider the following: The length of rhinoceros’ horns are normally distributed with a mean of
10.1 cm and a standard deviation of 1.4 cm.
14. On the graph to the right, label the mean and three
standard deviations above and below the mean.
15. What percent of rhinos have horns shorter than
10.1 cm? 50% Longer than 10.1cm? 50%
16. Calculate the probability of a rhinos’ horn being longer
than 9 cm. Interpret your results.
9  10.1
z
 0.78
1.4
Probability for a z-score of -0.78 is 0.2148; so 1-0.2148 = 0.7852
The probability that a rhino’s horn will be bigger than 9 cm is 78.52%.
17. Find P  9.5  x  10.5
z
10.5  10.1
 0.29  0.6141
1.4
z
9.5  10.1
 0.43  0.3336
1.4
0.6141-0.3336 = 0.2805 (28.05%)