probability staistics

8/17/2019 Probability Staistics

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Assignment of Probability and Statistics-DEC2013

Group No. 5(SEM-2)

Name Roll No.

Ujjwal Bopche 30 - Group LeaderVinita Nayak 34Shrutika Dawkhar 33Neha Shaha 32Mitali Purarkar 31 - Absent Pankaj Gurav 29 - Absent

Rahul Yadav 35 - Absent


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Q1(a)I- state and prove Bays theorem of probabilty.


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Ans-

Q1(a)II

Q1(b)


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Q2(a)i-


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Q4(a)ii-


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Q4(a) i] The mean weekly sales of chocolate bar in candy stores was 146.3

bars per store. After an advertising campaign the mean weekly sales in 22

stores for a typical week increased to 153.7 and showed standard deviation

of 17.2. Was the advertising campaign successful? [Given the table value of tat 5% level of significance for 21 degrees of freedom in 1.721]

Soln:Let us take the hypothesis that the advertising is not successful. Here is usual notations, we

are given,

X = 153.7, µ= 146.3 ,n = 22,δ =17.2

t= µ

√−

= 5.−.. √ 21 =1.92

No of degrees of freedom = 22-1= 21

Since the calculated value of t is greater than the critical value at 5% leve of significance. i.e

the diffrence significant at 5% level Hence we can say that the advertising is successful.


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ii] The number of accidents of taxi drivers in a year follows poisson

distribution with mean 5. Out of 500 such drivers find approximately the no.

of drivers with no accidents in a year.

Soln.Let X denote numberof accident in a year involving taxi drivers,

Givenmeanλ=3,

P(no.ofaccidentinayear)=P(X=0)

=

! =−

= −5 =0.00674

No.oftaxidriverswithnoaccidents =1000× 0.00674

=6.74 = 7(approx)

Q4(b) i] Calculate the co-efficient of variation from the following data:

Age Under (yrs) 10 20 30 40 50 60 70 80

No of persons 15 30 53 75 100 110 115 125

Soln.

Since Upper Limit of the class intervals and the commutative frequencies are given in this

prob, first the series will be changed onto the class intervals with ordinary frequencies and

then the coefficient of variation will be calculated.

Calculation of coefficient of variation

Age in yrs Mid

values (m)

No of persons Step

deviation(−5

)

d’x

Ƒd’x Ƒd’x2

0-10 5 15 -4 -60 240

10-20 15 15 -3 -45 135

20-30 25 23 -2 -46 92

30-40 35 22 -1 -22 22

40-50 45 25 0 0 0

50-60 55 10 1 10 10

60-70 65 5 2 10 20

70-80 75 10 3 30 90

N=125

∑▒ƒ ^′ = 123

∑ ƒ

= 609


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For the calculation of the coefficient of variation the mean and the standard deviation of the

series will have to be found out.

Calculation of mean :

a = × + (∑ ƒ

∗ )

=45 + (−5 ∗ 10 )

= 45 +9.84

=35.16 years

Calculation of standard deviation:

Ϭ = ∑ ƒď (∑ ƒď

)^2 * CϬ = 5 −5

* 10

Principles and practice of statistics,

=4.872 (98) * 10=√ 4.8729604 * 10=√ 3.911 * 10= 1.978 * 10

=19.78 years

Calculation of the coeffieient of variation :

C.V=Ϭ * 100

= .5. * 100= 5626 * 100

= 56.26%

ii] Find the probabililty of constructing a two digit even no. using the digits

1,2,3,4,5,6,7,8,9 if repetation of digits is not allowed.

Soln,

Since we want two digit even numbers with no repetition, we are seeking for numbers which end with 2/4/6/8.

Case 1: Numbers ending with 2/4/6/8. We now have 4 options to choose from and put at the unit's place.Let say we choose some digit (say 2) and put it in the unit's place.

Thus total number of combinations ending with 2 numbers = 8 * 1 = 8


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Similarly for othersThus total number of even combinations without repeatition of digits = (4*8) = 32 Q5(b)i

Q5(b)ii-


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s

Q6(a)i


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Q6(b)


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Q7(a)ii


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Q7(b)i

probability staistics

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