STATISTICS AND PROBABILITY DIPLOMA SYLLABUS HINT QUESTION
NEW PAST QUESTION FROM(2079-2070) SOLVED
Question 1:
Calculate
the correlation coefficient between ages of husbands and ages of wives for the
following bivariate frequency distribution:
(similar question ask in 2077/2076/2075/2073 )
|
Ages of husbands |
Ages of wives |
TOTAL |
||||
|
10-20 |
20-30 |
30-40 |
40-50 |
50-60 |
||
|
15-25 |
6 |
3 |
- |
- |
- |
9 |
|
25-35 |
3 |
16 |
10 |
- |
- |
29 |
|
35-45 |
- |
10 |
15 |
7 |
- |
32 |
|
45-55 |
- |
- |
7 |
10 |
4 |
21 |
|
55-65 |
- |
- |
- |
4 |
5 |
9 |
|
TOTAL |
9 |
29 |
32 |
21 |
9 |
100 |
FULL AND ENLARGE IMAGE: CLICK HERE
ANWSER:
Let, dy = (y − 35)/10, where assumed mean Ay
= 35 and y = 10.
dx = (x −
40) /10, where assumed mean Ax = 40 and h
= 10
|
CI |
10-20 |
20-30 |
30-40 |
40-50 |
50-60 |
|
|||||
|
MV(Y) |
15 |
25 |
35 |
45 |
55 |
|
|||||
|
CI |
MV(X) |
dy dx |
-2 |
-1 |
0 |
+1 |
+2 |
fx |
dx .fx |
dx 2.fx |
fxy.dxdy |
|
|
15-25 |
20 |
-2 |
4 6 24 |
2 3 6 |
- |
- |
- |
9 |
-18 |
-36 |
30 |
|
|
25-35 |
30 |
-1 |
2 3 6 |
1 16 16 |
0 10
0 |
- |
- |
29 |
-29 |
29 |
22 |
|
|
35-45 |
40 |
0 |
- |
0 10
0 |
0 15
0 |
0 7
0 |
- |
32 |
0 |
0 |
0 |
|
|
45-55 |
50 |
+1 |
- |
- |
0 7
0 |
1 10 10 |
2 4 8 |
21 |
21 |
21 |
18 |
|
|
55-65 |
60 |
+2 |
- |
- |
- |
2 4 8 |
4 5 20 |
9 |
18 |
36 |
28 |
|
|
fy |
9 |
29 |
32 |
21 |
9 |
N=100 |
Σdx .fx
=-8 |
Σdx 2.fx =122 |
Σfxy.dxdy =98 |
||
|
dy.fy |
-18 |
-29 |
0 |
21 |
18 |
Σdy.fy = -8 |
|
||||
|
dy 2.fy |
36 |
29 |
0 |
21 |
36 |
Σ dy 2.fy =122 |
|
||||
|
fxy.dxdy |
30 |
22 |
0 |
18 |
28 |
Σfxy.dxdy =98 |
|
||||
FULL AND ENLARGE IMAGE: CLICK HERE
Question 2: VERY IMPORTANT:
Find the regression equation of y on x and x on y from the following data.
[ Estimate y when x=10 (question ask in 2077) tail question ]
[ Estimate the Value X ,when Y=12 (question ask in 2075) tail question ]
[ Estimate y when x=125 (question ask in 2077) tail question ]
[ Estimate y(milleage) ,when x(Driving speed )=67 (question ask in 2073) tail question ]
|
X |
5 |
9 |
13 |
17 |
21 |
|
Y |
3 |
8 |
13 |
18 |
23 |
Estimate y when x=125
Estimate y(milleage) ,when x(Driving speed )=67
From
the following data of rainfall and production of rice,
find
the most likely production corresponding
to the rainfall of 80 mm
Production (quintals)
Rainfall (mm)
Mean 80
64
S.D. 3 4
Correlation=-0.4
Answer:
Let X and Y represent the production and rainfall respectively.
The Production(Quintals) =75.2
Calculate Geometric Mean for 14 , 36, 45 ,70 , 105
(similar question ask in 2078-79)
Solution:
Based on the given data, we have:
|
x |
logx |
|
14 |
1.1461 |
|
36 |
1.5563 |
|
45 |
1.6532 |
|
70 |
1.8450 |
|
105 |
2.0211 |
|
Total |
8.2217 |
Based on the above mentioned formula, Geometric
Mean G.M. will be:
The Geometric Mean of the given numbers is 44.09
And
Calculate Harmonic Mean for data 14 ,36 , 45 , 70 ,105
Solution:
Based on the given data, we have:
|
x |
1/x |
|
14 |
0.7142 |
|
36 |
0.2777 |
|
45 |
0.0222 |
|
70 |
0.0142 |
|
105 |
0.0095 |
|
Total |
1.0378 |
Based on the above mentioned formula, Harmonic
Mean H.M will be:
Calculate mean (A.M) ,median and mode from the following
data.
|
Wages(RS) |
0-10 |
10-20 |
20-30 |
30-40 |
40-50 |
50-60 |
|
No of Workers |
4 |
6 |
20 |
10 |
7 |
3 |
%20Negative%20skewness;%20b)%20Normal%20curve;%20c)%20Positive%20skewness%20(Durkhure...%20_%20Downl.png)
A bag
contains 8 white and 3 red balls. If two balls are drawn at random, find the
probability that (i) both are white (ii) both are red (iii) one is of each
color.
ANSWER:
Total balls =11
M=possible outcomes =Number of ways of drawing 2 balls from 11 balls = 11c2
=55 Ways
N=Number of ways of drawing both White balls =8c2=28 ways
K=Number
of ways of drawing both red balls =3c2=3ways
L=Number
of ways of drawing one of each color.=
(selection
of 1 out 8 white balls) and(selection of 1 out of 3 red balls)
8c1 *3c1 =8*3=24ways
i) P(BOTH ARE WHITE)= N/M =28/55
ii) P(BOTH ARE RED) =K/M=3/55
iii)P(ONE IS OF EACH COLOR)=L/M=24/55
A class
consists of 60 boys and 40 girls. It two students are chosen at random, What
will be the probability that
i) Both
are Boys
ii) One
boys and One girl.
ANSWER:
Total persons =40+60=100
M=possible outcomes
=Number of ways of choosing 2 student
from 100 students
= 100c2
=4950 Ways
N=Number
of ways of choosing both of Boys.
i.e
(selection of 2 out of 60 boys)
=60c2=1770 Ways
L= Number
of ways of Choosing One boys and girl.
= selection
of 1 out of 60 boys and selection of 1 out of 40 girl
=60c1 *40c1 =2400 Ways
i) P(BOTH ARE BOYS)=
N/M =1770/4950
ii) P(ONE BOY AND ONE GIRL
) =L/M=2400/4950
A bag contains 4 white and 5 red balls and 6 black balls. If two
balls are drawn at random, find the probability that (i) both
are of same Colour (ii) one is white. (iii) both are of Different Colours.
ANSWER:
Total
balls =4+5+6=15
M=possible outcomes
=Number of ways of drawing 2 balls from 15
balls
= 15c2
=105Ways
N=Number of ways of drawing both of same Colour
i.e (selection of 2 out of 4 white balls) or( selection of 2 out
of 5 red balls) or (selection of 2 out of 6 black balls) =4c2+5c2+6c2=6+10+15
=31 ways
K= Number of ways of drawing one is white
i.e [ selection of 1 out 4
white balls and selection of 1 out of 5 red balls ] Or [selection
of 1 out 4 white balls and selection of 1 out of 6black balls ]
=4c1 *5c1+4c1 *6c1=4*5+4*6=44 Ways
L= Number of ways of drawing both are of Different Colours
=[selection of 1 out 4 white balls and selection of 1 out of 5 red balls] or [selection of 1 out of 5
red balls and selection of 1 out of
6 Black balls] or [selection of 1 out 4 white balls and selection of 1 out of 6
Black balls]
=4c1 *5c1+5c1 *6c1+4c1 *6c1
=4*5+5*6+4*6=74ways
i) P(BOTH ARE OF SAME
COLOUR)= N/M =31/105
ii) P(ONE IS OF WHITE)
=K/M=44/105
iii)P(BOTH ARE OF DIFFERENT
COLOR)=L/M=74/105
If A,B and C are three mutually
exclusive events with:
FIND P(A) , P(B) AND P(C) ?
Soln,
Mutually
exclusive is a statistical term describing two or more events that
cannot happen simultaneously
i.eP(AnB)=P(BnC)=P(AnC)=P(AnBnC)=0
( FIRST METHOD
sum of all the probabilities is 1
P(A)+P(B)+P(C)=1 )
For finding
the condition P(A)+P(B)+P(C)=1
You can use
any of it.
( Next
method Or,P(AuBuC)= P(A)+P(B)+P(C)- P(AnB)-P(BnC)-P(AnC)+P(AnBnC)
Or, P(AuBuC)=
P(A)+P(B)+P(C)
Or, P(A)+P(B)+P(C)=1 )
Equating
first and second,
2P(A)=P(B)
Equating
second and third,
P(c)=4P(B)
P(A)+P(B)+P(C)=1
Or, P(A)+P(B)+4P(B)=1
Or, 1/2 P(A)+5P(B)=1
.png){kind=small}
Mean of random
variable: If X is the random
variable and P is the respective probabilities, the mean of a random variable
is defined by:
Mean (Expected value) = ∑ XP
where variable X
consists of all possible values and P consist of respective probabilities.
Variance of Random
Variable: The variance tells how
much is the spread of random variable X around the mean value. The formula for
the variance of a random variable is given by;
Var(X) = σ2 = E(X2) – [E(X)]2
where E(X2) = ∑X2P and E(X) = ∑ XP
Important
relation:
👉 Mean E(X) = ∑ X.P(x)
👉Varience(X)
= σ2 = E(X2) – [E(X)]2 = ∑X2P-∑ X.P(x)
👉The sum of all the possible probabilities is
Σ P(x)=1
Find the mean(Expected value),variance and S.D of the discrete random variable X whose probability distribution is
|
X |
-2 |
1 |
2 |
3.5 |
|
P(X) |
0.21 |
0.34 |
0.24 |
0.21 |
Solution
Using the definition of
mean gives mean summation(x.P(x))
Mean(expected value)
E(X) =∑x.P(x)=(−2)(0.21)+(1)(0.34)+(2)(0.24)+(3.5)(0.21)=1.135
Mean =1.135
E(X2)= ∑x2.P=(−2)2(0.21)+(1)2(0.34)+(2)2(0.24)+(3.5)2(0.21)=4.7125
Variance σ2 = E(X2)
– [E(X)]2 =4.7125-1.135=3.5775
S.D =√ σ2 =√3.5775=1.8914
Question1:
same question ask in 2076/2071
A random variable X has
the following probability distribution
|
X |
-2 |
-1 |
0 |
1 |
2 |
3 |
|
P(X) |
0.1 |
k |
0.2 |
2k |
0.3 |
K |
Find the value of K, expected
value of X, variance
We know,sum of all the possible probabilities is
Σ P(x)=1
Or, 0.1+k+0.2+2k+0.3+k=1
Or, 4K=0.4
K=0.1
So, table becomes:
|
X |
-2 |
-1 |
0 |
1 |
2 |
3 |
|
P(X) |
0.1 |
0.1 |
0.2 |
0.2 |
0.3 |
0.1 |
Mean(expected
value)
E(X) =∑x.P(x)=(−2)(0.1)+(-1)(0.1)+(0)(0.2)+(1)(0.2)+(2)(0.3)+(3)(0.1)=0.8
For variance,
E(X2)= ∑x2.P=(−2)2(0.1)+(-1)2(0.1)+(0)2(0.2)+(1)2(0.2)+(2)2(0.3)+(3)2(0.1) =2.8
Variance σ2 = E(X2)
– [E(X)]2 =2.8-0.8=2
Binomial Distribution
The discrete
probability distribution derived from the Bernoully process
is known as Binomial
distribution.
The
The following are three basic assumption under which
Binomial
distribution can be used:
1.
Random experiment should be performed for the fixed number of times.
2.
The experiment should have only two
events known as "success"
and
“failure”
3.
All the experiments performed should be independent
of one another.
The
probability of success denoted by p should be constant per every
experiment.
When the probability
of a success in one trial is known, the probabilities
of success of exactly
once, twice, thrice, .. etc. in n trials can also be
known.
Notations for Binomial Distribution and the Mass Formula:
Where:
§ P is the probability
of success on any trail.
§ q = 1- P – the probability
of failure
§ n – the number of
trails/experiments
§ x = the number of
successes, it can take the values 0, 1, 2, 3, . . . n.
§ nCx = n!/x!(n-x) and denotes the number of combinations of n elements
taken x at a time.
Assuming
what the nCx means,
we can write the above formula in this way:
Just to
remind that the ! symbol after a number means
it’s a factorial. The factorial of a non-negative integer x is denoted by
x!. And x! is the product of all positive integers less than or equal to x. For
example, 4! = 4 x 3 x 2 x 1 = 24.
Question 1:
same question ask in 2078-79
The probability of hitting a target is1/5 or20%.
If six hitting are made, find the probability that (i) none will strike the
target (ii) exactly 2 will strike the target
Solution:
Question 2:
same question ask in 2077
A
A dice is thrown 3 times. Getting an
even number is considered as a success.
Find the probability of getting
i)No success
ii) At least one Success
Solution:
n=3
P= probability of getting an even number on one trail(Getting
an even number(2,4,6) is considered as a success).
==3/6=1/2
q = 1- P – the probability of failure( not Getting an even number assumed
as failure)
=1-1/2
=1/2
i) P(x=0) = 6C0 P0. Q6=1*(1/2)0.(1/2)6=1/64
ii) P(x>=1) = P(1)+p(2)+p(3)
=1-[p(0)] { Σ P(x)=1 }
=1-1/64
=63/64
Question 3:
same question ask in 2075
The average percentage of a failure in a certain examination
is 25%. What is the probability that out of 5 students
2 or more students will pass the examination?
SOLUTION:
Question 4:
same question ask in 2076
The
incidence of occupations diseases in an industry is such that the workman have
a 20% chance of suffering from it. What is the probability that out of six men,
four or more will contact the decreases?
SOLUTION:
n=6
P= probability of
success on any trail(i.e chance of suffering assumed as success).
=20%=1/5
q = 1- P – the probability of failure(chance of not suffering
assumed as failure)
=1-1/5
=4/5
i.e four or more will contact the decreases i.e x>=4
P(x>=4)=p(4)+p(5)+p(6)
P(4) = 6C4 P4. q2=15*(1/5)4.(4/5)2=48/3125
P(5) = 6C5 P5. q= 6*(1/5)5. (4/5)=24/15625
P(6) = 6C6 P6. q0=1 *(1/5)6. (4/5)0=1/15625
P(x>=4)=p(4)+p(5)+p(6)
=48/3125+24/15625+1/15625
=53/3125
Hence,probability that four or more will contact the decreases=53/3125
CONCEPT OF
Mean and standard deviation of
Binomial distribution
If p be the probability of a success and q
that of a failure in one trial, then the probabilities of 0, 1, 2, 3,... .. n
successes in n trials are listed above which are the successive terms of the
binomial expansion of (q + p)n . Hence the distribution is known as
binomial distribution. The mean and the standard deviation of the binomial
distribution are np and √(npq) respectively. The two independent constants n
and p (or q) are known as the parameters.
Question 1:
same question ask in 2077
Poisson distribution
Poisson distribution
is actually another probability distribution formula. As per binomial distribution,
we won’t be given the number of trials or the probability of success on a
certain trail. The average number of successes will be given in a certain time
interval. The average number of successes is called “Lambda” and denoted by the
symbol “λ”.
The formula for Poisson Distribution formula is given below:
Mean and Variance of
Poisson Distribution
If λ is the average number of successes occurring in a given time
interval or region in the Poisson distribution, then the mean and the variance
of the Poisson distribution are both equal to λ.
E(X) = λ
and
V(X) = σ2 = λ
Question 1:
same question ask in 2078-79
On average there are
four road accidents per week in Kathmandu Using Poisson Distribution, find the
probability of:
(a) no accident per week
(b) 3 accidents per week
(c) at least 2 accidents per week.
Wehave,
P(X)= (λ -x .e-Ʌ )/x!
Where, λ=Average number or mean
x=possion random variable
e=2.7182(approx)
λ=4
i)P(X=0)=
(4-0 e-4)/0!
=1/ e4=1.831*10-2
ii)P(X=3)= (4-3 e-4)/3! =1/( e4 *6*64 )= 4.769*10-5
iii)P(X<2)=P(0)+P(1)
P(X=1)= (4-1 e-4)/1!
=1/ e4 *4= 4.578 *10-3
iii)P(X>=2)=
P(2)+P(3)….
=1-[ P(0)+P(1) ]
Σ P(x)=1
=1-[ 1/
e4+1/ e4 *4 ]
=0.9771
Question 2:
same question ask in 2073
If average number of road accident per month at certain intersection of road is 4 By using the Poisson Probability
Find i) Probability of no accident in a month ii) Probability of 2 accidents in a month. iii) probability of less than 2 accidents in a month
Solution :
λ=4
Wehave,
P(X)= (λ -x .e-Ʌ )/x!
Where, λ=Average number or mean
x=possion random variable
e=2.7182(approx)
i)P(X=0)= (4-0 e-4)/0! =1/ e4=1.831*10-2
ii)P(X=2)= (4-2 e-4)/2! =1/( e4 *2*16 )= 5.723*10-4
iii)P(X<2)=P(0)+P(1)
P(X=1)= (4-1 e-4)/1! =1/ e4 *4= 4.578 *10-3
iii)P(X<2)=P(0)+P(1)
=1/ e4+1/ e4 *4
=2.28*10-2
Example.1
