Desklib - Online Library for Study Material with Solved Assignments, Essays, Dissertations

Desklib is an online library for study material with solved assignments, essays, dissertations, etc. It offers a wide range of study materials for various subjects and courses. Desklib provides solutions to academic problems.

Running Head: QUANTITATIVE METHODS
Quantitative Methods
Name of the student:
Name of the university:
Course ID:

QUANTITATIVE METHODS 1
Table of Contents
Answer 1..........................................................................................................................................2
Answer 1. a).................................................................................................................................2
Answer 1. b).................................................................................................................................2
Answer 1. c).................................................................................................................................2
Answer 1. d).................................................................................................................................2
Answer 2..........................................................................................................................................3
Answer 2. a).................................................................................................................................3
Answer 2. a. i)..........................................................................................................................3
Answer 2. a. ii).........................................................................................................................3
Answer 2. a. iii)........................................................................................................................4
Answer 2. b).................................................................................................................................4
Answer 3..........................................................................................................................................4
Answer 4..........................................................................................................................................5
Answer 4. a).................................................................................................................................5
Answer 4. b).................................................................................................................................5
Answer 4. c).................................................................................................................................5
Answer 4. d).................................................................................................................................5
Answer 4. e).................................................................................................................................6
Answer 4. f)..................................................................................................................................6
Answer 5..........................................................................................................................................6
Answer 6..........................................................................................................................................7
Answer 6. a).................................................................................................................................7
Answer 6. b).................................................................................................................................7
Answer 7..........................................................................................................................................7
Answer 7. a).................................................................................................................................7
Answer 7. b).................................................................................................................................8

2QUANTITATIVE METHODS
Answer 1.
Answer 1. a)
n=20 , p=0.5 .
P (x = 12) = n!
x ! ( n−x ) ! px (1− p)n− x
= 20!
12! ( 20−12 ) ! (0.5)12 (1−0.5)20−12
= 20 !
12! 8 ! (0.5)12 (0.5)8
= 125970*(0.5)20 = 0.1201.
Answer 1. b)
n=20 , p=0.3 .
P (x > 8) = 1−P ( x ≤ 8 ) =1−∑
x=0 ( n
x ) px(1− p)n−x
= 1−∑
x=0 ( 20
8 ) (0.3)8 (1−0.3)20−8
= 1−∑
k=0 (20
8 )(0.3)8
(0.7)12
= 1 – 0.8866 = 0.1133.
Answer 1. c)
n=20 , p=0.7 .
P (x < 12) = P ( x ≤ 12 )−P ( x=12 )
¿ [ ∑
x=0 ( n
x ) px ( 1− p ) n−x
]- [ n !
x ! ( n−x ) ! px (1− p)n−x ]
= ∑
x=0 (20
12 )(0.7)12 (1−0.7)20−12− 20!
12! ( 20−12 ) ! (0.7)12(1−0.7)20−12
= ∑
k=0 ( 20
12 ) (0.7)12
(0.3)8− 20 !
12! 8 ! (0.7)12(0.3)8
= 0.2277 – 0.1143 = 0.1133.
Answer 1. d)

3QUANTITATIVE METHODS
n=15 , p=0.4 .
P (4 ≤ x ≤ 12) = P ( x ≤ 12 )−P ( x< 4 )
P ( x ≤ 12 )
¿ [ ∑
x=0 ( n
x ) px ( 1− p ) n−x
]
= ∑
x=0 ( 15
12 ) (0.4)12 (1−0.4)15−12
= ∑
k=0 (15
12 )(0.4)12
(0.6)3
= 0.9997
P (x < 4) = P ( x ≤ 4 ) −P ( x=4 )
¿ [ ∑
x=0 ( n
x ) px ( 1− p ) n−x
]- [ n !
x ! ( n−x ) ! px (1− p)n−x ]
= ∑
x=0 ( 15
4 ) (0.4)4 (1−0.4 )15−4 − 15!
4 ! ( 15−4 ) ! (0.4)4 (1−0.4)15−4
= ∑
k=0 (15
4 )(0.4)
4
(0.6)11− 15 !
4 ! 11 ! (0.4)4 (0.6)11
= 0.2173 – 0.1268 = 0.0905.
P (4 ≤ x ≤ 12) = P ( x ≤ 12 )−P ( x< 4 )= 0.9997 - 0.0905 = 0.9092.
Answer 2.
30% (p = 0.3) students are enrolled in marketing.
18% (p = 0.18) students are enrolled in international business.
20 students (n=20) are chosen at random.

4QUANTITATIVE METHODS
Answer 2. a)
Answer 2. a. i)
The probability that at least half of the students (p = 0.5) are studying a marketing major
= P
(Zi ≤ 0.3−0.5
√ 0.5∗ ( 1−0.5 )
20 )=P ¿) = P ( Zi ≤ −0.2
0.111803 ¿=P( Zi ≤−1.78885)
= (1-0.9632) = 0.0368.
Answer 2. a. ii)
The probability that no more than a quarter of the students (p = 0.75) are studying an
international business major =
P
(Zi ≥ 0.18−0.75
√ 0.75∗ ( 1−0.75 )
20 )=P ¿) = P (Zi ≥ −0.57
0.096825 ¿=P( Zi ≥−5.88693)
= (1- 1.967*10-9).
Answer 2. a. iii)
The probability that the number of students studying a marketing major is between 10 (p
= 10
20 = 0.5) and 15 (p = 15
20 = 0.75) is given as-
P( 0.3−0.75
√ 0.75∗( 1−0.25 )
20
≤ Zi ≤ 0.3−0.5
√ 0.5∗( 1−0.5 )
20
)
= P ( −0.45
0.096825 ≤ Zi ≤ −0.2
0.1118 ¿=P(−4.64758≤ Zi ≤−1.78885)
= P ( Zi ≤−1.78885 ) −P ( Zi ≤−4.64758 ) = (0.03682 - 1.679* 10-6) = 0.036818.
Answer 2. b)
The mean of the number of students studying marketing major = n*p = 20*0.3 = 6.
The standard deviation of the student studying marketing major = n*p*(1-p) = 20*0.3*(1-
0.3) = 20*0.3*0.7 = 4.2 ≈ 4.

5QUANTITATIVE METHODS
The total proportion of students studying marketing major or international business major
is ( ^p)= (0.3+0.18) = 0.48.
The mean of the number of students studying marketing major = n* ^p=20∗0.48=9.6 ≈ 10
.
The standard deviation of the number of students studying marketing major = n* ^p*(1- ^p)
= 20*0.48*(1-0.48) = 20*0.48*0.52 = 4.99 ≈ 5.
Answer 3.
The weights of eggs are normally distributed with average 55 gm. and standard deviation
1 gm.
The empirical rule of normality defines that the 95% confidence intervals are estimated
by two standard deviations. The mean of eggs lies in the interval of –
((55 gm. - 2*1 gm.), (55 gm. + 2*1 gm.)) = ((55-2) gm., (55+2) gm.) = (53 gm., 57 gm.).
Therefore, a likely chosen random egg from the farm would have the following weight of
53 gm.
Answer 4.
Answer 4. a)
μ=604 , σ =56.8 , x ≤635
The calculated z-statistic = Zi = x−μ
σ = 635−604
56.8 = 31
56.8 =¿ 0.54577464788.
The probability for the normal distribution = P (
x ≤ 635 ¿=P ( x−μ
σ ≤ 635−604
56.8 )=P(Zi ≤ 0.54577464788)=0.7074 .
Answer 4. b)
μ=48 , σ =12, x <20
The calculated z-statistic = Zi = x−μ
σ = 48−20
12 = 28
12 =¿ 2.3333.

6QUANTITATIVE METHODS
The probability for the normal distribution = P (
x <48 ¿=P( x−μ
σ < 20−48
12 )=P(Zi ≤−2.3333)=0.009815.
Answer 4. c)
μ=111, σ =33.8 ,100 ≤ x< 150
The probability for the normal distribution =
P (100 ≤ x <150 ¿=P (x<150)−P(x<100)
¿ P ( x−μ
σ <150−111
33.8 )−P ( x−μ
σ < 100−111
33.8 )
¿ P(Zi ≤ 1.15384615385)−P (Zi ≤−0.32544378698)=(0.8757−0.3724)=0.5033 .
Answer 4. d)
μ=264 , σ=10.9, 250<x <255
The probability for the normal distribution =
P (250< x <255 ¿=P (x< 255)−P( x<250)
¿ P ( x−μ
σ <255−264
10.9 )−P ( x−μ
σ < 250−264
10.9 )
¿ P(Zi ≤−0.82568807339)−P (Zi ≤−1.28440366972)=(0.2045−0.0995)=0.105 .
Answer 4. e)
μ=37 , σ=4.35 , x >35
The calculated z-statistic = Zi = x−μ
σ = 35−37
4.35 = −2
4.35 =¿ -0.45977011494.

7QUANTITATIVE METHODS
The probability for the normal distribution = P (
x >35 ¿=P ( x−μ
σ > 35−37
4.35 )=P ( Zi >−0.45977011494 )=1−P ( Zi ≤−0.45977011494 ) =1−0.3228=0.6772 .
Answer 4. f)
μ=156 , σ=114 , x ≥170
The calculated z-statistic = Zi = x−μ
σ = 170−156
114 = 14
114 =¿ 0.12280701754.
The probability for the normal distribution = P (
x ≥ 35 ¿=P ( x −μ
σ ≥ 170−156
114 )=P ( Zi ≥ 0.12280701754 ) =1−P ( Zi<0.12280701754 )=1−0.5489=0.4511.
Answer 5.
The age of real-estate investors is normally distributed that has mean (μ) = 40 years and
standard deviation = 10 years.
The proportion of investors who are below 25 years old-
= P ( x <25 ¿ = P ( x−μ
σ < 25−40
10 ¿=¿P (Zi < −15
10 ¿=P ( Zi ←1.5 ) =0.06681
Hence, the age of 6.681% of the real-estate investors whose age is normally distributed
are less than 25 years old.
Answer 6.
The mean of home mortgage in New Zealand = $283000
The standard deviation of home mortgage in New Zealand = $50000
The home mortgages in New Zealand is normally distributed.
Answer 6. a)

8QUANTITATIVE METHODS
The proportion of home loans that is more than $250000 is given as-
P (x >250000 ¿ = P ( x−μ
σ > 250000−283000
50000 ¿=¿P ( Zi >−33000
50000 ¿=1−P ( Zi ≤−0.66 )
¿ 1−0.2546=0.7454 .
Answer 6. b)
The proportion of home loans that are between $250000 and $300000 =
P (250000< x <300000 ¿ = P (x < 300000) – P (x ≤ 250000)
= P ( x−μ
σ < 300000−283000
50000 )−¿P ( x−μ
σ < 250000−283000
50000 ¿=¿P (
Zi < 17000
50000 ¿−P( Zi< −33000
50000 )
¿ P ( Zi ≤0.34 )- P( Zi ≤−0.66)
¿ 0.6331−0.2546=0.3785 .
Answer 7.
Answer 7. a)
The standard deviation = 12.56.
71.97% of the values are greater than 56.
P ¿) = 0.7197
Or, Zi = -0.582.
Or, x−μ
σ =¿ -0.582
Or, x−μ
σ =¿ -0.582

9QUANTITATIVE METHODS
Or, 56−μ
12.56 =¿ -0.582
Or, 56 – μ = (-0.582) * (12.56) = -7.30992
Or, μ = 56 + 7.30992 = 63.30992 ≈ 63.31
Therefore, the value of μ is 63.31.
Answer 7. b)
The mean = 352.
Only 13.35% of the values are less than 300.
P ¿) = 0.1335
Or, Zi =−1.11
Or, 300−352
σ = -1.11
Or, −52
σ =−1.11
Or, σ = −52
−1.11 =46.84685
Hence, the value of σ is 46.84685.