statistics examples 3

54 56 56 59 60 62 62 66 67 6868 70 70 73 73 73 75 77 79 7979 81 81 82 83 83 85 86 86 8889 89 90 90 91 93 93 94 95 98

Assignment 3

Answer to the question no - 21

Maximum = 98 Minimum = 54

Let us determine the number of classes with the help of the following formula:

No of Classes: 2K ≥ N

2K ≥ 40

log 2K ≥log 40

k log 2 ≥ log 40

k ≥log 40log 2

K ≥ 5.32 ∴ K ≈ 5

Class Interval = Upper Limited−Lower Limited

K = 100−505 = 10

Now we will take 10 as the class interval and the first class as 50-60 and Mid point of a class will be:

Mid Points = upper limit of the class+lower limit of the class

2

The computation are set out in the table:

Classes Tally Freq. (f i) R.F. C.F. C.R.F.

50-60 5 5/ 40 5 5/ 4060-70 8 8 /40 13 13/ 4070-80 8 8 /40 21 21/4080-90 13 13/ 40 34 34 /40

90-100 6 6 /40 40 40 /40Sums=40 1

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N = 40k = Number of Classes

Upper limit = 100Lower limit = 50

Assignment 3


4.7 4.7 4.0 4.7 3.0 4.4 5.0 3.3 3.8 6.4

3.3 3.6 4.7 4.4 5.4 3.0 4.9 5.2 4.24.1 6.0 5.8 4.9 3.8

Maximum = 6.4 Minimum = 3.0Let us determine the number of classes with the help of the following formula: No of Classes: 2K ≥ N

2K ≥ 25

log 2K ≥log 25

k log 2 ≥ log 25

k ≥log25log2

K ≥ 4.64 ∴ K ≥ 5


K = 7.0−2.05 = 1

Now we will take 1 as the class interval and the first class as 50-60 and Mid point of a class will be:


2



2.0-3.0 2 2/25 2 2/253.0-4.0 7 7 /25 9 9 /254.0-5.0 11 11 /25 20 20/255.0-6.0 4 4 /25 24 24 /256.0-7.0 1 1/25 25 25/25

Sums=25 1

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Assignment 3


38.0 24.5 21.5 30.8 20.3 24.0 29.6 19.4 25.6 39.513.3 28.0 30.8 32.9 30.3 19.9 24.6 32.3 24.7 18.736.8 31.2 50.9 30.7 30.3

Maximum = 50.9 Minimum = 13.3

Let us determine the number of classes with the help of the following formula:

No of Classes: 2K ≥ N

2K ≥ 25

log 2K ≥log 25

k log 2 ≥ log 25

k ≥log25log2

K ≥ 4.64 ∴ K ≥ 5


K = 52.0−12.05 = 8

Now we will take 8 as the class interval and the first class as 12.0-20.0 and Mid point of a class will be:


2



12.0-20.0 4 4 /25 4 4 /2520.0-28.0 9 9 /25 13 13/2528.0-36.0 8 8 /25 21 21/2536.0-44.0 3 3/25 34 34 /2544.0-52.0 1 1/25 25 25/25

Sums=25 1

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Upper limit = 52.0Lower limit = 12.0

Assignment 3



From the table we have,

Population Mean: μx=∑i=1

k

f imi

N

¿f 1m1+ f 2m2+−−−−−∓f kmk

N

¿ 13540 = 3.375 (Mean)

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Rating (mi) Number of Students (f i)

C.F. f imi

1 1 1 12 7 8 143 15 23 454 10 33 405 7 40 35

Sums = 40 Sums = 135

Five Number SummeryMin 1Q1 3

Q2/Median 3Q3 4

MAX 5

∑i=1

k

f imi=135

∑i=1

k

f i=N=40

Assignment 3

Median:

MED =

=

=

20 TH observation : From the cumulative frequencies we see that the 20TH observation is the 12TH value at the rating 3. ∴ 20TH observation = 3

21 ST observation : From the cumulative frequencies we see that the 21ST observation is the 13TH value at the rating 3. ∴ 21ST observation = 3

Median ¿ 20TH+21ST

2=3+3

2 = 3 (Median)

Interquartile Range (IQR): Since there are N = 40 observations,

IQR = Q3 - Q1

Q1 = .25(N+1)TH = .25(40+1)TH = 10.25TH

So, the first quartile is one quarter of the way from the 10TH observation to the 11TH.

10 TH observation : From the cumulative frequencies we see that the 10TH observation is the 2nd value at the rating 3. ∴ 10TH observation = 3

11 TH observation : From the cumulative frequencies we see that the 11TH observation is the 3rd value at the rating 3. ∴ 11TH observation = 3

J = 6, f = 8, L = 60, U = 70

∴ Q1 = 3 + .25(3 – 3) = 3

Now, Q3 = .75(N+1)TH = .75(40+1)TH = 30.75TH

So, the third quartile is three quarters of the way from the 30TH observation to the 31st.

∴ 30 TH observation : From the cumulative frequencies we see that the 30TH observation is the 7TH value at the rating 4. ∴ 30TH observation = 4

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[ N2 ]TH

+[ N+22 ]

TH

2

[402 ]

TH

+[40+22 ]

TH

2

20TH+21ST

2

Assignment 3

∴ 31 ST observation : From the cumulative frequencies we see that the 30TH observation is the 8TH value at the rating 4. ∴ 31ST observation = 4

∴ Q3 = 4+ .75(4 – 4) = 4

∴ IQR = Q3 – Q1 = 4 – 3 = 1

Variance: σ x2

=

∑i=1

k=5

f imi2

N−μx

2

=

49940

−3 . 382

=12.48-11.42 =1.06

Standard Deviation: σ =√σ2⇒σ =√1.06⇒σ =1.03




Sample Mean: X=

∑i=1

k

f imi

N

¿f 1m1+ f 2m2+−−−−−∓f kmk

N

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Number of claims (mi) Number of policies (f i)

C.F. f imi

0 21 21 01 13 34 132 5 39 103 4 43 124 2 45 85 3 48 156 2 50 12

Sums = 50 Sums = 70

Five Number SummeryMin 0Q1 0

Q2/Median 1Q3 2

MAX 6

∑i=1

k

f imi=70∑i=1

k

f i=n=50

Assignment 3

¿ 7050 = 1.4 (Mean)

Median:

MED = =

=

25 TH observation : From the cumulative frequencies we see that the 25TH observation is the 4TH value at the number of claims 1. ∴ 25TH observation = 1


Median ¿ 25TH+26TH

2=1+1

2 = 1 (Median)

Interquartile Range (IQR): Since there are n = 50 observations,

IQR = Q3 - Q1

Q1 = .25(N+1)TH = .25(50+1)TH = 12.75TH

So, the first quartile is three quarters of the way from the 12TH observation to the 13TH.



∴ Q1 = 0 + .75(0 – 0) = 0

Now, Q3 = .75(N+1)TH = .75(50+1)TH = 38.25TH

So, the third quartile is one quarter of the way from the 38TH observation to the 39TH.

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[ n2 ]TH

+[ n+22 ]

TH

2

[502 ]

TH

+[50+22 ]

TH

2

25TH+26TH

2

Assignment 3

∴ 38 TH observation : From the cumulative frequencies we see that the 38TH observation is the 4TH value at the number of claims 2. ∴ 38TH observation = 2

∴ 39 TH observation : From the cumulative frequencies we see that the 39TH observation is the 5TH value at the number of claims 2. ∴ 39TH observation = 2

∴ Q3 = 2 + .25(2 – 2) = 2

∴ IQR = Q3 – Q1 = 2 – 0 = 2

Variance: σ x2

=

∑i=1

k=5

f imi2

N−μx

2

=

24850

−1. 42

=4.96-1.96=3

Standard Deviation: σ =√σ2⇒σ =√3⇒σ =1.73

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Assignment 3



Study Time (Hours)

Number of Students (f i) C.F. mi f imi

0-4 3 3 2 64-8 7 10 6 42

8-12 8 18 10 8012-16 5 23 14 7016-20 2 25 18 36


Mid Points: mi=Upper Limit of the class+Lower limit of the class

2


Sample Mean: X=

∑i=1

k

f imi

N

¿f 1m1+ f 2m2+−−−−−∓f kmk

N

¿ 23425 = 9.36 (Mean)

Median: MED =

= = 13TH

13 TH observation : From the cumulative frequencies we see that the 13TH observation is the 3rd value in the class 8-12, then

J = 3, f = 8, L = 8, U = 12

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Five Number SummeryMin 0Q1 5.70

Q2/Median 9.25Q3 12.8

MAX 20

[n+12 ]

TH

∑i=1

k

f i=n=25 ∑i=1

k

f imi=234

[25+12 ]

TH

Assignment 3

∴ 13TH observation = L+(J−12 )U−L

f=8+(3−1

2 ) 12−88

=9.25

Median ¿13TH observation = 9.25


IQR = Q3 - Q1

Q1 = .25(N+1)TH = .25(25+1)TH = 6.5TH

So, the first quartile is half of the way from the 6TH observation to the 7TH.

6 TH observation : From the cumulative frequencies we see that the 6TH observation is the 3rd value in the class 4-8, then

J = 3, f = 7, L = 4, U = 8

∴ 6TH observation ¿ L+(J−12 )U−L

f=4+(3−1

2 )8−47

=5.42

7 TH observation : From the cumulative frequencies we see that the 7TH observation is the 4TH value in the class 4-8, then

J = 4, f = 7, L = 4, U = 8


f=4+(4−1

2 ) 8−47

=5.99

∴ Q1 = 5.42 + .5(5.99 – 5.42) = 5.70

Now, Q3 = .75(N+1)TH = .75(25+1)TH = 19.5TH

So, the third quartile is half of the way from the 19TH observation to the 20TH.

∴ 19 TH observation : From the cumulative frequencies we see that the 19TH observation is the 1ST value in the class 12-16, then

J = 1, f = 5, L = 12, U = 16


f=12+(1−1

2 ) 16−125

=12.4

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Assignment 3

∴ 20 TH observation : From the cumulative frequencies we see that the 20TH observation is the 2nd value in the class 12-16, then

J = 2, f = 5, L = 12, U = 16


f=12+(2−1

2 ) 16−125

=13.2

∴ Q3 = 12.4 + .5(13.2 – 12.4) = 12.8

∴ IQR = Q3 – Q1 = 12.8 – 5.70 = 7.1

Variance: σ x2

=

∑i=1

k=5

f imi2

N−μx

2

=

3∗22+7∗62+8∗102+5∗142+2∗182

25−9.362

=107.68-87.61=20.07

Standard Deviation: σ =√σ2⇒σ =√20 .07⇒σ =4.48

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Assignment 3



Study Time (Hours)


0-5 4 3 2.5 7.55-10 7 10 7.5 52.5

10-15 8 18 12.5 10015-20 4 22 17.5 7020-25 2 24 22.5 45



2


Sample Mean: X=

∑i=1

k

f imi

N

¿f 1m1+ f 2m2+−−−−−∓f kmk

N

¿ 27525 = 11.46 (Mean)

Median: MED =

= =

12 TH observation : From the cumulative frequencies we see that the 12TH observation is the 2nd value in the class 10-15, then

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Five Number SummeryMin 0Q1

Q2/Median 9.25Q3

MAX 25

∑i=1

k

f imi=275∑i=1

k

f i=n=24

12th+13 th

2

[242 ]

th

+[24+22 ]

th

2

[ n2 ]th

+[n+22 ]

th

2

Assignment 3

J = 2, f = 8, L = 10, U = 15


f=8+(3−1

2 ) 12−88

=9.25

Median ¿13TH observation = 9.25

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Assignment 3



Study Time (Hours)


0-10 29 29 5 14510-20 23 52 15 34520-30 17 69 25 42530-40 14 83 35 49040-50 11 94 45 49550-60 6 100 55 330

Sums = 100 Sums = 2230


2


Sample Mean: X=

∑i=1

k

f imi

N

¿f 1m1+ f 2m2+−−−−−∓f kmk

N

¿ 2230100 = 22.30 (Mean)

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Five Number SummeryMin 0Q1 8.51

Q2/Median 19.12Q3 34.46

MAX 60

∑i=1

k

f imi=2230∑i=1

k

f i=n=100

Assignment 3

Median: MED =

=

= 50TH+51TH

2

50 TH observation : From the cumulative frequencies we see that the 50TH observation is the 21ST value in the class 10-20, then

J = 21, f = 23, L = 10, U = 20


f=10+(21−1

2 ) 20−1023

=18.91


J = 22, f = 23, L = 10, U = 20


f=10+(22−1

2 ) 20−1023

=19.34

Median ¿ 50TH+51TH

2 =

18.91+19.342 = 19.12


IQR = Q3 - Q1

Q1 = .25(n+1)TH = .25(100+1)TH = 25.25TH

So, the first quartile is one quarter of the way from the 25TH observation to the 26TH.


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[ n2 ]TH

+[ n+22 ]

TH

2

[1002 ]

TH

+[100+22 ]

TH

2

Assignment 3

J = 25, f = 29, L = 0, U = 10


f=0+(25−1

2 ) 10−029

=8.44


J = 26, f = 29, L = 0, U = 10


f=0+(26−1

2 ) 10−029

=8.73

∴ Q1 = 8.44 + .25(8.73 – 8.44) = 8.51

Now, Q3 = .75(n+1)TH = .75(100+1)TH = 75.75TH

So, the third quartile is three quarters of the way from the 75TH observation to the 76TH.

∴ 75 TH observation : From the cumulative frequencies we see that the 75TH observation is the 6TH value in the class 30-40, then

J = 6, f = 14, L = 30, U = 40


f=30+(6−1

2 ) 40−3014

=33.9

∴ 76 TH observation : From the cumulative frequencies we see that the 76TH observation is the 7TH value in the class 30-40, then

J = 7, f = 14, L = 30, U = 40


f=30+(7−1

2 ) 40−3014

=34.64

∴ Q3 = 33.92 + .75(34.64 – 33.92) = 34.46

∴ IQR = Q3 – Q1 = 34.46 – 8.51 = 25.95

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statistics examples 3

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