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Data | Frequency | Relative Frequency | Cumulative Relative Frequency |
---|---|---|---|
33 | 1 | 0.032 | 0.032 |
42 | 1 | 0.032 | 0.064 |
49 | 2 | 0.065 | 0.129 |
53 | 1 | 0.032 | 0.161 |
55 | 2 | 0.065 | 0.226 |
61 | 1 | 0.032 | 0.258 |
63 | 1 | 0.032 | 0.29 |
67 | 1 | 0.032 | 0.322 |
68 | 2 | 0.065 | 0.387 |
69 | 2 | 0.065 | 0.452 |
72 | 1 | 0.032 | 0.484 |
73 | 1 | 0.032 | 0.516 |
74 | 1 | 0.032 | 0.548 |
78 | 1 | 0.032 | 0.580 |
80 | 1 | 0.032 | 0.612 |
83 | 1 | 0.032 | 0.644 |
88 | 3 | 0.097 | 0.741 |
90 | 1 | 0.032 | 0.773 |
92 | 1 | 0.032 | 0.805 |
94 | 4 | 0.129 | 0.934 |
96 | 1 | 0.032 | 0.966 |
100 | 1 | 0.032 | 0.998 (Why isn't this value 1? ANSWER: Rounding) |
Recall that for grouped data we do not know individual data values, so we cannot describe the typical value of the data with precision. In other words, we cannot find the exact mean, median, or mode. We can, however, determine the best estimate of the measures of center by finding the mean of the grouped data with the formula:
$Mean\text{}of\text{}Frequency\text{}Table=\frac{{\displaystyle \sum fm}}{{\displaystyle \sum f}}$
where
$f=$ interval frequencies and
m = interval midpoints.
Just as we could not find the exact mean, neither can we find the exact standard deviation. Remember that standard deviation describes numerically the expected deviation a data value has from the mean. In simple English, the standard deviation allows us to compare how “unusual” individual data is compared to the mean.
Find the standard deviation for the data in [link] .
Class | Frequency, f | Midpoint, m | $f*m$ | $f(m-\stackrel{\u2013}{x}{)}^{2}$ |
---|---|---|---|---|
0–2 | 1 | 1 | $1*1=1$ | $1(1-7.58{)}^{2}=43.26$ |
3–5 | 6 | 4 | $6*4=24$ | $6(4-7.58{)}^{2}=76.77$ |
6-8 | 10 | 7 | $10*7=70$ | $10(7-7.58{)}^{2}=3.33$ |
9-11 | 7 | 10 | $7*10=70$ | $7(10-7.58{)}^{2}=41.10$ |
12-14 | 0 | 13 | $0*13=0$ | $0(13-7.58{)}^{2}=0$ |
26=n | $\stackrel{\u2013}{x}=\frac{197}{26}=7.58$ | ${s}^{2}=\frac{306.35}{26-1}=12.25$ |
For this data set, we have the mean,
$\stackrel{\u2013}{x}$ = 7.58 and the standard deviation,
s
_{x} = 3.5. This means that a randomly selected data value would be expected to be 3.5 units from the mean. If we look at the first class, we see that the class midpoint is equal to one. This is almost two full standard deviations from the mean since 7.58 – 3.5 – 3.5 = 0.58. While the formula for calculating the standard deviation is not complicated,
${s}_{x}=\sqrt{\frac{\Sigma {(m-\stackrel{\u2013}{x})}^{2}f}{n-1}}$ where
s
_{x} = sample standard deviation,
$\stackrel{\u2013}{x}$ = sample mean, the calculations are tedious. It is usually best to use technology when performing the calculations.
The standard deviation is useful when comparing data values that come from different data sets. If the data sets have different means and standard deviations, then comparing the data values directly can be misleading.
#ofSTDEVs is often called a " z -score"; we can use the symbol z . In symbols, the formulas become:
Sample | $x$ = $\overline{x}$ + zs | $z=\frac{x\text{}-\text{}\stackrel{\u2013}{x}}{s}$ |
Population | $x$ = $\mu $ + zσ | $z=\frac{x\text{}-\text{}\mu}{\sigma}$ |
Two students, John and Ali, from different high schools, wanted to find out who had the highest GPA when compared to his school. Which student had the highest GPA when compared to his school?
Student | GPA | School Mean GPA | School Standard Deviation |
---|---|---|---|
John | 2.85 | 3.0 | 0.7 |
Ali | 77 | 80 | 10 |
For each student, determine how many standard deviations (#ofSTDEVs) his GPA is away from the average, for his school. Pay careful attention to signs when comparing and interpreting the answer.
$z=\#\; of\; STDEVs=\frac{\text{value}\u2013\text{mean}}{\text{standarddeviation}}=\frac{x-\mu}{\sigma}$
For John, $z=\#ofSTDEVs=\frac{2.85\u20133.0}{0.7}=\u20130.21$
For Ali, $z=\#ofSTDEVs=\frac{77-80}{10}=-0.3$
John has the better GPA when compared to his school because his GPA is 0.21 standard deviations below his school's mean while Ali's GPA is 0.3 standard deviations below his school's mean.
John's z -score of –0.21 is higher than Ali's z -score of –0.3. For GPA, higher values are better, so we conclude that John has the better GPA when compared to his school.
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