Chapter 9: Section 9-3 Statistics - Creighton University · · 2015-07-16Chapter 9: Section 9-3 Statistics D. S. Malik Creighton University, Omaha, NE D. S. Malik Creighton University,

Chapter 9: Section 9-3Statistics

D. S. MalikCreighton University, Omaha, NE

D. S. Malik Creighton University, Omaha, NE () Chapter 9: Section 9-3 Statistics 1 / 36

Definition(Mean) Let x1, x2, . . . , xn be n real numbers. The mean (or average) ofthese numbers, written x , is the real number defined by

x =x1 + x2 + · · ·+ xn

n.

ExampleThe midterm test scores of a class of 10 students are 87, 76, 97, 56, 66,87, 99, 45, 88, 97. Then the mean of the midterm test scores is

x =87+ 76+ 97+ 56+ 66+ 87+ 99+ 45+ 88+ 97

10

=79710

= 79.7


Empty Slide for Notes


ExampleAt a fitness center new exercise bikes are becoming very popular. On aparticular weekend, the fitness center logged the time, in minutes, aparticular machine was used by 40 different members and the followingdata was collected:

35 45 10 30 30 20 15 60 45 38 5 5 15 2890 35 32 5 56 15 10 20 25 15 34 70 80 6032 56 45 30 15 45 30 15 56 89 35 38

x =

35+ 45+ 10+ 30+ 30+ 20+ 15+ 60+ 45+ 38+ 5+ 5+15+ 28+ 90+ 35+ 32+ 5+ 15+ 56+ 10+ 20+ 25+ 15+34+ 70+ 80+ 60+ 32+ 56+ 45+ 30+ 15+ 45+ 30+15+ 56+ 89+ 35+ 38

40

=141440

= 35.35.

The average minutes a member used one of the new bikes is 35.35minutes.D. S. Malik Creighton University, Omaha, NE () Chapter 9: Section 9-3 Statistics 4 / 36

x Frequency (f ) xf5 3 5 · 3 = 1510 2 10 · 2 = 2015 6 15 · 6 = 9020 2 20 · 2 = 4025 1 25 · 1 = 2528 1 28 · 1 = 2830 4 30 · 2 = 12032 2 32 · 3 = 6434 1 34 · 1 = 3435 3 35 · 3 = 10538 2 38 · 2 = 7645 4 45 · 4 = 18056 3 56 · 3 = 16860 2 60 · 2 = 120


x Frequency (f ) xf70 1 70 · 1 = 7080 1 80 · 1 = 8089 1 89 · 1 = 8990 1 90 · 1 = 90

Sum ∑ f = 40 ∑ xf = 1414

x =∑ xf∑ f

=141440

= 35.35.




Remark(i) The symbol ∑ is known as the summation symbol. This symbol isread as “summation”.(ii) ∑ f is read as the sum of the values of f .

Definition (Mean of grouped data)Suppose that a data set consists of the numbers x1, x2, . . . , xk such thatthe number x1 appears f1 times, x2 appears f2 times, . . . , xk appears fktimes. (In general, the number xi appears fi times, i = 1, 2, . . . , k.) Thenthe mean (average) of the numbers is given by

x =x1f1 + x2f2 + · · ·+ xk fkf1 + f2 + · · ·+ fk

=∑ki=1 xi fi

∑ki=1 fi

.


ExampleThe heights (in inches) of students in a class are given in the following table:

Height (x) Frequency (f )63 464 665 566 267 368 569 470 371 572 273 274 375 1


ExampleNext, we multiply each height with its frequency, and then find the totalfrequency and the total sum of the values as follows:

Height (x) Frequency (f ) xf63 4 63 · 4 = 25264 6 64 · 6 = 38465 5 65 · 5 = 32566 2 66 · 2 = 13267 3 67 · 3 = 20168 5 68 · 5 = 34069 4 69 · 4 = 27670 3 70 · 3 = 21071 5 71 · 5 = 35572 2 72 · 2 = 14473 2 73 · 2 = 14674 3 74 · 3 = 22275 1 75 · 1 = 75

∑ f = 45 ∑ xf = 3062


Example Continued

ExampleThus, we have, ∑ f = 45, ∑ xf = 3062. Hence,

x =∑ xf∑ f

=306245≈ 68.

Hence, the mean or average of the heights is 68 inches = 5 feet 8 inches.




Example(Mean of grouped data in interval form) A local music store wants to knowthe average amount a customer spends in the store. After the holiday season theylooked at their sales data and the following data was retrieved.

Amount in dollars Frequency (f )0− 6 67− 13 1014− 20 921− 27 1228− 34 2035− 41 1842− 48 1549− 55 856− 62 1063− 69 570− 76 2


Example Continued

ExampleFirst we find the midpoint of each interval and then assume that the number oftimes the midpoint of the interval appears in the data set is equal to thefrequency of the numbers in that interval.

Amount in dollars Midpoint (x) Frequency (f )0− 6 3 67− 13 10 1014− 20 17 921− 27 24 1228− 34 31 2035− 41 38 1842− 48 45 1549− 55 52 856− 62 59 1063− 69 66 570− 76 73 2


Example Continued

ExampleNext, we multiply each height with its frequency, and then find the totalfrequency and the total sum of the values as follows:

Amount in dollars Midpoint (x) Frequency (f ) xf0− 6 3 6 187− 13 10 10 10014− 20 17 9 15321− 27 24 12 28828− 34 31 20 62035− 41 38 18 68442− 48 45 15 67549− 55 52 8 41656− 62 59 10 59063− 69 66 5 33070− 76 73 2 146

∑ f = 115 ∑ xf = 4020


Example Continued

ExampleThus, we have, ∑ f = 115, ∑ xf = 4020. Hence,

x =∑ xf∑ f

=4020115

≈ 34.96.

Hence, the average money a customer spent in the store is $34.96.


Median

DefinitionLet x1, x2, . . . , xn, be the numbers in a data such that x1 ≤ x2 ≤ · · · ≤ xn.The median of these numbers is defined as follows:(i) Suppose that n is odd, say n = 2k + 1, where k is an integer. Then themedian of the data set is

x n+12= xk+1.

(Note that n+12 = 2k+1+12 = k + 1.)

(ii) Suppose that n is even, say n = 2k, where k is an integer. Then themedian is the number

x n2+ x n

2+1

2=xk + xk+1

2.


ExampleConsider the following data:

16, 15, 20, 6, 28, 10, 34, 2, 25, 8, 19.

IThis data in ascending order is:

2, 6, 8, 10, 15, 16, 19, 20, 25, 28, 34.

This data set has 11 numbers. The median is at position (1+ 11)/2 = 6,that is, median is:

median = 16


ExampleConsider the following data set:

4, 10, 14, 18, 20, 22, 28, 36, 39, 40.

This data set has 10 entries. It has no (single) middle element. So, welook at the 5th and 6th entries. (Note that (1+ 10)/2 = 5.5, and wehave no element at position 5.5. So we look at the elements in the 5th and6th positions.)

In this data set, the elements in the 5th and 6th positions are 20 and 22,respectively. Next we add these numbers and divide by 2, i.e., we take theaverage of these two numbers. Now

20+ 222

= 21.

Hence, the median of the data 4, 10, 14, 18, 20, 22, 28, 36, 39, 40 is 21.




Example(Finding the median of grouped data). Consider the following groupeddata:

x Frequency (f )4 37 513 220 725 532 336 638 142 2

Let us find the median of this data set.


ExampleThe numbers in the data set are already in increasing order. Next we findthe total number of elements in this data set by adding their frequencies.Now

3+ 5+ 2+ 7+ 5+ 3+ 6+ 1+ 2 = 34.

Thus, the number of elements is 34 = 2 · 17, an even number of elements.So we look at the elements in the 17th and 18th positions.Note that first 4 appears three times, then 7 appears five times, next 13appears twice, and so on. Note that the 17th element is 20 and the 18thelement is 25. Hence, the median is

20+ 252

= 42.5.




Example(Finding the median of grouped data in interval form) Consider thefollowing grouped data in interval form.

Interval Frequency (f )0− 8 29− 17 418− 26 627− 35 336− 44 145− 53 1

First we find the midpoint of each interval and then assume that thenumber of times the midpoint of the interval appears in the data set isequal to the frequency of the numbers in that interval.


Example

Interval x Frequency (f )0− 8 4 29− 17 13 418− 26 22 627− 35 31 336− 44 40 145− 53 49 1

The total number of elements in this data is:2+ 4+ 6+ 3+ 1+ 1 = 17.Thus, the number of elements is 17, i.e., anodd number of elements. So we look at the element at the 17+1

2 = 9thposition. From the frequency column it follows that the number at the 9thposition is 22. Hence, the median is 22.




Mode

DefinitionSuppose that we have a set of numbers x1, x2, . . . , xn. Suppose that themaximum number of times a number appears in this set is m. Then thenumber(s) xi that appears m times in this set are called the mode(s) ofthese numbers.


ExampleConsider the following set of numbers.

3.1, 2.5, 1.2, 2.7, 1.2, 2.5, 1.8, 0.9, 2.5, 3.2, 0.5, 2.6

Let us make a frequency table of these numbers. We have

x Frequency (f )0.5 10.9 11.2 21.8 12.5 3

x Frequency (f )2.6 12.7 13.1 13.2 1

This implies that the maximum number of times a number appears in thisdata is 3. Because 2.5 is the only number that appears 3 times, the modeof this data is 2.5.




ExampleConsider the following set of numbers.

5, 14, 5, 35, 20, 5, 10, 14, 5, 14, 14, 7, 10, 27, 2, 10, 35, 10

Let us make a frequency table of these numbers. We have

x Frequency (f )2 15 47 110 4

x Frequency (f )14 420 127 135 2

This implies that the maximum number of times a number appears in thisdata set is 4. Because each of the numbers 5, 10, and 14 appear 4 times,the modes of this data set are 5, 10, and 14.


Range

DefinitionSuppose that we have a set of numbers x1, x2, . . . , xn. Suppose that xi isthe smallest number and xj is the largest number in this set. Then xj − xiis called the range of these numbers.

ExampleConsider the set of numbers

35, 20, 56, 5, 48, 30, 35, 48, 50, 20, 18, 34, 27.

The smallest number in this set is 5 and the largest number is 56. So therange of these numbers is

56− 5 = 51.




Standard Deviation

DefinitionLet x1, x2, . . . , xn be n real numbers. Let x be the mean of these numbers.(i) The variance of these numbers, written Var , is the real number

Var =(x1 − x)2 + (x2 − x)2 + · · ·+ (xn − x)2

n− 1 .

(ii) The standard deviation, written s, of these numbers is the realnumber

s =√Var =

√(x1 − x)2 + (x2 − x)2 + · · ·+ (xn − x)2

n− 1

=

√(x21 + x

22 + · · ·+ x2n )− nx2

n− 1 .


ExampleConsider the following set of numbers: 10, 8, 2, 4, 6. The mean of thesenumbers is:

x =10+ 8+ 2+ 4+ 6

5=305= 6

Nowx x − x (x− x)22 2− 6 = −4 (−4)2 = 164 4− 6 = −2 (−2)2 = 46 6− 6 = 0 (0)2 = 08 8− 6 = 2 (2)2 = 410 10− 6 = 4 (4)2 = 16

Then n = 5 and ∑(x − x)2 = 16+ 4+ 0+ 4+ 16 = 40.Thus,

Var =∑(x − x)2n− 1 =

405− 1 =

404= 10 and s =

√Var =

√10 ≈ 3.16.






Documents

Chapter 9: Section 9-3 Statistics - Creighton University · · 2015-07-16Chapter 9: Section 9-3 Statistics D. S. Malik Creighton University, Omaha, NE D. S. Malik Creighton University,