statistic : descriptive measures

STATISTIC : DESCRIPTIVE MEASURES

Measures For Central TendencyMean, Mode, Median ungrouped data

Mean _x =

∑x

nWhere n is the sample

Mean

µ = ∑x

NWhere N is the population

Example : Mean

The number of 911 calls classified as domestic disturbance calls in large metropolitan location where sampled for 10 randomly selected 24 hour periods with the following results. Find the mean number of calls per 24 hours period

10 20 50 30 20 40 20 30 10 20

Mean _x =

∑250

10= 25



Median

Median of a set of data is a value that divides the bottom 50% of the data from the top 50% of the data. To find the median of a data set, first arrange the data inincreasing order. If the number of observations is odd then the median is the Number in the middle of the observation list. If the number is even then the medianIs the mean of the two values closest to the middle of the ordered list

~x~µ= Sample median = Population median

Example : Median

The number of 911 calls classified as domestic disturbance calls in large metropolitan location where sampled for 10 randomly selected 24 hour periods with the following results. Find the median number of calls per 24 hours period

10 10 20 20 20 20 30 30 40 50 (even number of observation)

20 + 20~x =2

= 20


Mode The mode is the value in a data set that occurs the most often. If no such valueexists, we say that the data has no mode. If two such values exist, we say thedata is bimodal. If three such values exist we say the data set is trimodal. There is no symbol that is used to represent the mode.

Data set : 10, 12, 15, 15, 18, 20 Mode: 15

Shapes of Distribution

(i) Bell-shaped(ii) Left-skewed(iii) Right-skewed

Bell-shaped


Shapes of Distribution

Left-skewed

Right-skewed


Measures For Central TendencyMean, Mode, Median grouped data

Age Frequency Class Marks Class width

5-14 7 9.5 10

15-24 15 19.5 10

25-34 5 29.5 10

35-44 5 39.5 10

45-54 5 49.5 10

Mean = x =∑xf n = where x represent the class marks, f represent the

frequencies and n represent the sample size

Mean = x = 9.5 x 7 + 19.5 x 15 + 29.5 x 5 + 39.5 x 5 + 49.5 x 5 37

Mode

Spread evenly the number of size, size is 37 so the middle value should be 19 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37

Median size


5-14 7 9.5 10

15-24 15 19.5 10

25-34 5 29.5 10

35-44 5 39.5 10

45-54 5 49.5 10

According to the table the median size would be at the class range 15-24.To be able to reach the median size we need to add 12 to the upper frequency Values from the above class range ( to reach 19) so median is calculated as

Median = The lower boundary of class range + 12/15 x class size= 14.5 + (12/15) x 10)


5-14 7 9.5 10

15-24 15 19.5 10

25-34 5 29.5 10

35-44 5 39.5 10

45-54 5 49.5 10

Mode

The modal class is defined to be the class with maximum frequency. TheMode for grouped data would be the class mark of the modal class.

Mode = 19.5


Range, Variance, and Standard DeviationUngrouped Data

Range

The range for a data set is equal to the maximum value in the data set minusthe minimum value in the data set.

Example :

Range in test score for Johan and Jamal. Range for Johan is 100 – 85 = 15and the range for Jamal is 90 – 60 = 30. The spread in Johan’s score is asmeasure by range twice the spread of Jamal’s score.



Variance

s2 = ∑( x – x )2

n-1

_

Variance for sample of size n

δ2 = ∑( x – µ )2

N

_

Variance for population of size N



Variance: continue…

Example : Times in minutes for 5 students to complete a task were5, 10, 15, 3 and 7. The mean time is 8 minutes (refer back to mean). See Table3.0, it illustrates the computation indicated by the formula variance forsample.

Score Deviation from mean (x – x ) Squares of deviations (x – x )2

5 5 – 8 = -3 -32

10 10 – 8 = 2 22

15 15 – 8 = 7 72

3 3 – 8 = -5 -52

7 7 – 8 = -1 -12

∑(x – x )=0 (sums of deviations) ∑(x – x )2=88 (sums of squares of deviations)

_ _

_ _



∑( x – x )2

n-1

_88 4

Variance: continue…

If we followed the variance formula in the previous slide, the variance for Table 3.0is 22 minutes squared. s2 = = = 22

The standard deviation is then calculated as

s = √s2 Sample standard deviation

δ =√δ2 Population standard deviation

The standard deviation is √22 = 4.7 minutes


Coefficient Of Variation

The coefficient variation is equal to the standard deviation divided by the mean.The result is usually multiplied by 100 to express it as a percent. The coefficient of variation for a sample is given by

CV = x 100%

The coefficient of variation for population

CV = x 100%

sx

µδ

Example :

A national sampling of prices for new and used cars found that the meanprice for a new car is $20,000 and the standard deviation is $6,125 and thatthe mean price for a used car is $5,485 with a standard deviation equal to $2,730. In terms of absolute variation, the standard deviation of price for newcars is more than twice that of used cars. However, in terms of relative variation, there is more relative variation in the price of used cars than in new cars.

The CV for used cars is and the CV for new cars is2,7305,485

X 100 = 49.8%


Coefficient Of Variation : continue…

6,12520,100

X 100 = 30.5%

Exercise : Answer all Question

Q1 . Table below gives the selling prices in tens of thousands of dollars for 20Homes sold during the past month. Find the mean, mode, and median.

60.5 113.5 79.0 475.5

75.0 70.0 122.5 150.0

100.0 125.5 90.0 175.5

89.0 130.0 111.5 100.0

50.0 340.5 100.0 525.0

Mean = Mode= Median=


20-29 11

30-39 25

40-49 14

50-59 7

60-69 3

Q2 . Find the mean, mode, and median for the grouped data below

Mean = Mode= Median=

Q3. Fill in the table below with the details required

Color Car No involved in accident

Deviation from mean (x – x ) Squares of deviations (x – x )2

Red 10

Blue 5

Yellow 10

Green 8

Purple 7

∑(x – x )= ∑(x – x )2=

_

_

_

_

Variance s2 = =∑( x – x )2

n-1

_

statistic : descriptive measures

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