http://www.ck12.org Chapter 6. Normal Distribution Curves
x ̄=15 + 65 + 55 + 35 + 45 + 25
6
=
240
6
= 40
s=√
∑(x−x ̄)^2
n− 1s=√
625 + 625 + 225 + 25 + 25 + 225
5
s=√
1 , 750
5
=
√
350 ≈ 18. 7
Now let’s calculate the standard deviation of the data for Brand B had it been a sample:
x ̄=40 + 50 + 35 + 40 + 45 + 30
6
=
240
6
= 40
s=√
∑(x−x ̄)^2
n− 1s=√
0 + 100 + 25 + 0 + 25 + 100
5
s=√
250
5
=
√
50 ≈ 7. 1
Notice that the standard deviation of the data for Brand A (18.7) is still much larger than the standard deviation of
the data for Brand B (7.1).
Example C
Suppose data are normally distributed, with a mean of 84 and a standard deviation of 18. Between what 2 values
will the following proportions of the data fall?
a. 68%
b. 95%
c. 99.7%
a. 68% of the data will fall within 1 standard deviation of the mean. Therefore, 68% of the data will fall between
84 −18 and 84+18, or between 66 and 102.
b. 95% of the data will fall within 2 standard deviations of the mean. Therefore, 95% of the data will fall between
84 −( 2 × 18 )and 84+( 2 × 18 ), or between 48 and 120.
c. 99.7% of the data will fall within 3 standard deviations of the mean. Therefore, 99.7% of the data will fall between
84 −( 3 × 18 )and 84+( 3 × 18 ), or between 30 and 138.
Points to Consider
- Does the value of standard deviation stand alone, or can it be displayed with a normal distribution?
- Are there defined increments for how data spreads away from the mean?
- Can the standard deviation of a set of data be applied to real-world problems?