case 1- own problem

8/10/2019 Case 1- Own Problem

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JEEVANANTHAM. K(31059)

DEVAKUMARAN .C.K (31123)

ASSIGNMENT

Tamil Nadu has increasing number of recreation places. Among them parks are of high prominence,

basically parks are classified into childrenspark, lovers park, and theme park. The entrance fees for

the park varies based upon the table. 1 shows that range of entrance fee for the park and the number

of parks with particular fee, whereas Table. 2 shows the type of park and the number of that

particular park situated in the top cities like Chennai, Salem, and Trichy.

Park Entrance fees Parks with

particular fee

Childrens park

20-40 31

40-60 25

60-80 28

Lovers park

80-100 19

100-120 18

120-140 24

Theme park

140-160 17

160-180 11

180-200 27

Park Chennai Salem Trichy Total

Childrens park 28 19 37 84

Lovers park 17 31 13 61

Theme park 11 23 21 55

Total 56 73 71 200


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Entrance fees Frequency Mid pt Freq. midPt Cum. freq (m- )2 (m- )2.f

20-40 31 30 930 31 5256.25 157687.5

40-60 25 50 1250 56 2756.25 137812.5

60-80 28 70 1960 84 1056.25 73987.5

80-100 19 90 1710 103 156.25 14062.5

100-120 18 110 1980 121 56.25 6187.5

120-140 24 130 3120 145 756.25 98312.5

140-160 17 150 2550 162 2256.25 338437.5

160-180 11 170 1870 173 4556.25 774562.5

180-200 27 190 5130 200 7656.25 1454688

Total 200 990 20500 24506.25 3055688

=mean

fm/ f = 20500 / 200 = 102.5

Variance

2= (m- )2. f / n-1 = 15355.21

S.D = 123.91

Median

Q1 = 55.2 + 56 / 2 = 55.6

Q2 Median = 96.842 + 17.844 / 2 = 97.368

Q3 145.882 + 147.058 /2 = 146.47

Range

200- 20 = 180


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Interquartile range =Q3-Q1 =90.87

co-efficient of variation: (S.D/Mean) x 100 =120.88

Mode= 292.104 205 = 87.107

Skewers = (mean mode)/ S.D = (102.5 - 87.107) / 123.91

=0.1242

Q&A:

1. What is the probability of childrens park in Chennai?28/300 = 0.14

2. What is the probability of Lovers Park in Trichy?

13/200 =0.065

3. What is the probability of theme park is neither in Chennai nor in Salem?

21/200 =0.105

4. What is the probability of that there is a childrens park given that the city in

Chennai?

28/56 = 0.5

5. Event A is that there is a lovers park Event B is that there is a park in Chennai,

find whether both are independent?

Event A = 61/200 Event B = 56/200

P (A).P (B) = (61/200) x (56/200)

= 0.0854

P (A B) = 17 /200

= 0.065

Thus both are dependent event


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Joint probability:

Bayes Theorem:

Probability that a person in Rajus family will go to the park in given below,

find the posterior probability using bayes theorem.

0.4406

Binomial Distribution:

Consider 8 families independently are going for a week trip and the probability

of any family going to a childrens park is 0.6, theme park is 0.35 & Lovers Park is 0.05.

Q/A:


Childrens park 0.14 0.095 0.185 0.42

Lovers park 0.085 0.155 0.065 0.305

Theme park 0.055 0.115 0.105 0.275

Total 0.28 0.365 0.355 1


Childrens park 0.14 0.095 0.185 0.42

Loverspark 0.085 0.155 0.065 0.305

Theme park 0.055 0.115 0.105 0.275

Total 0.28 0.365 0.355 1


Childrens park 0.14 0.095 0.185 0.42

Lovers park 0.085 0.155 0.065 0.305

Theme park 0.055 0.115 0.105 0.275

Total 0.28 0.365 0.355 1

Park Prior

probability

Chances Joint

probability

Posterior

probability

Childrens park 0.43 0.42 0.1806 0.4098

Lovers park 0.51 0.35 0.1555 0.3529

Theme park 0.38 0.275 0.1045 0.2371


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1. What is the probability that exactly 4 families will meet in theme park?

2. What is the probability that none of the family will go to the lovers park?

3. What is the probability that at least 6 families will go to the Childrens Park?

Solution:n= 8

P childrens park = 0.6 q = 0.4

P Lovers Park = 0.05 q = 0.95

P theme park = 0.35 q = 0.65

1. 8c4 .(0.35)4 (0.65)4 = 70 x 0.001500 x 0.81450

= 0.855225.

2. 8c0 .(0.05)0 (0.95)8 = 1 x 1 x 0.663420

= 0.663420

3. [ 8c6 .(0.6)6 (0.4)2 + 8c7 .(0.6)7 (0.4)1 + 8c8 .(0.6)8 (0.4)0 ]

= (28 x 0.04665 x 0.16) + (8 x 0.02799 x 0.4) + (1 x 0.01679 x 1)

= 0.208992 + 0.089568 + 0.01679

=0.31535

Confidence interval & standard error:

A social welfare group wanted to study the pattern of lovers coming to Lovers Park.

It decided that the members of group will ask a single question to every lovers in that park

will u marry in future and if they get answer more than 70 % they will not take away

further step and if not they will publish on article in daily newspaper regarding this . On a

fine day they conducted this study and total of 200 couples were asked this question .the

mean number of couple who said yes are 138 with a S.D of 18.

1. Construct a 63% confidence interval.


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2. Construct a 95.5 % confidence interval and say whether they will publish an article

if they consider 95.5% confidence interval.

3. Calculate standard deviation for mean.

1. SE = / n

= 18 / 200

= 1.27

68.3 % confidence interval = 1SE

= 138 1.27

= 136.73, 139.27

2. 95.5% confidence interval = 2SE

= 138 2 (1.27)

= 138 2.54

= 135.46, 140.54

3. SE = / n = 1.27 ( thus the article wont be published in news paper )

Correlation & regression:

To determine correlation and regression between the peoples who are in childrens

park and the no of ice creams sales in that park.

Time No of visitor

in park (x)

ice cream

sales park (y)

(x-x) (y-) (x-x)2 (y-)2 (x-x). (y-)

10- 12 8 0 -20.6 -12 424.36 144 61107.84

12-2 12 10 -16.6 -2 275.56 4 1102.24

2-4 19 14 -9.6 2 92.16 4 368.64

4-6 32 16 3.4 4 11.56 16 184.96

6-8 72 20 43.4 8 1883.56 64 120547.84


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Correlation coefficient (rk) = (x-x). (y -) / {(x-x)2 . (y -)2}

=183311.52 / (2687.2 x 232)

= 0.29403

Regression line = y = mx + c

Where m = (x-x). (y -)/ (x-x)2

= 183311.52/ 2687.2

m= 68.2165

c= -m x= 12- 68.21x 28.6

= 12- 1950.8 = - 183.08

Thereby y= 68.21x- 183.08

Substitute the value ofx and to determine the new y value

X New (y) (y -) (y -)2.

8 362.6 350.6 122640.04

12 635.44 623.44 388677.43

19 111.91 1100.91 1212002.82

32 199.64 1987.64 3950712.77

72 472.04 4716.04 222241033.28

(y -)2= 227915.67

Find correlation of determination:

(rk) 2= (0.29403)2

=0.0864

case 1- own problem

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