Patrick B. answered • 08/21/21
Math and computer tutor/teacher
Set up a table like this:
X devX dev^2 X^2
59
63
68
70 -11.8 139.24 4900
...
93
97
99
------------------------------------
Totals
Per MS EXCEL...
The total of the stats is 1636.
The mean is 1636/20 = 81.8
The mode is 83, the most frequently occuring stat
The Median is 83, the average of the two middle stats is (83+83)/2 = 83
because the 10th stat and 11th stat are both 83.
The DEVIATION of the stat is the stat MINUS the mean.
For example the deviation of the stat 70 is 70-81.1 = -11.8.
The SUM of the deviations MUST be zero, and it is...
The sum of the deviations squared is 2095.2; that is squaring each deviation
THe sum of each stat squared is 135920. that is squaring each statistic.
The variance by definiton is sum of the deviations squared / (n-1)
2095.2/19 = 110.27368421052631578947368421053
The variance by shortcut formula is
(sum of the square of each stat - N*mean^2)/(n-1)
= (135920 - 20*81.8^2)/19 = 2095.2/19 = 110.27368421052631578947368421053
The standard deviation is the square root of the variance...
so 10.501127758985047020723208240463....
The first quartile Q1 is the 25th percentile. 25% of the data, or 5 stats
are at or below this value. There are many ways to choose the first quartile
that satisfy this defintion.
SInce there are 20 stats 25% of 20 = 5. So the 5th stat is 77.
Therefore Q1=77 as 5 stats, or 25% of the data is at or below this value.
Excel says the 1st quartile is 78.5, but not sure how that is defined without
reading the MS Excel documentation. Realistically, anything between 77 and 79 will
do since 25% of the data, or 5 stats are less than or equal to these values.
Likewise the 3rd quartile Q3 is the 75th percentile.
Since there are 20 stats 75% of 20 = 15. The 15th stat is
87, so Q3=87. That is, 25% of the data, or 5 stats are
at or ABOVE this value. MS Excel says that Q3=87.5.
realistically, any value betwen 87 and 88 will suffice.
the z-score formula is (x- mean)/stndDev
So the z-score for x=92 is (92 - 81.8)/10.501127758985047020723208240463...
= 0.97132424574804415455143077367407
so the score of x=92 is LESS than 1 standard devation from the mean.
This is not unusual, and in fact, per the Emprical rule, 2/3 of the data
is within one standard deviation of the mean in anyway. So no, it is not unusual.
range = max-min = 99-59 = 40. 5 classes are required for the frequency table, so
the width of each class shall be 40/5 = 8
class frequency
55-63 2
64-72 2
73-81 4
82-90 8
91-99 4
You can also do:
class freq
59-67 2
68-76 2
77-85 9
86-94 5
95-102 2
the range of (mean-1*stndDev, mean+1*stndeDev) = (71.29, 92.3)
there are 13 of the 20 stats in that range which is 65% of the data
the range of (mean - 2*stndDev, mean + 2 * stndDev) = (60.8, 102.8)
there are 19 of the 20 stats in that range which is 95% of the data
the empirical rule is supported by the data. Therefore we have a
normally distributed data set which is skewed, since the mean is 81.8
and the median is 83.
The skew estimate is (83-81.8)/20 = 1.2/20 = 0.06 = 6%
59 60 . . . . . . . . . . . . . . . . 77 . . . .81.8 . . 83 . . . 87. . . . . . . . . . . 99
q1 mean median q3
David B.
Or use R language. It does well with plots as well as the basic descriptives. The following script will provide the necessary data, , descriptives, histogram and box plot. ------------------------------------ V1 = c(77, 59, 84, 83, 70, 80 ,83, 82, 85, 92, 97, 88, 87, 86, 99, 93, 79, 63, 81, 68) summary(V1) table(V1) # catagory with most entries is mode V = var(V1) sigma = sd(V1) Z92 = (92-mean(V1))/sigma Pz = pnorm(Z92, lower.tail=FALSE) hist(V1, breaks = 5 ) boxplot(V1) ---------------------------- 1st Qu. Median Mean 3rd Qu. Mode Var sd Z(92) P(Z>.97) 78.50 83.00 81.80 87.25 82 110.27 10.50 .97. .17 Answer to 8: there is a 17% chance of getting a 92 or greater score, this is not particularly unusual.08/23/21