Essay About P-Cf And U-L

Quantitative Methods
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CONTENT
Question 1
Question 2
Question 3
Question 4
Question 5
BIBLIOGRAPHY
QUESTION 1
Class Interval of iPods
Freq f
Class Midpt, x
20-50
44100
50-80
608400
80-110
17728900
110-140
250000
140-170
384400
=3440
=19015800
1.2.1
1.2.2
1.2.3.
1.3.1
= (k/100)(n)
= (65/100)(40)
= 26th position
65th percentile
= L+[(P-cf)/f (U-L)]
= 80.5+[(26-18)/14 (110.5-80.5)]
= 80.5 + 17.14
= 97.14 or 97
1.3.2
Q1 = 1(N)th value
= (1(40))/4 th value
= 10th value
10th value is in the interval 50.5 – 80.5
Therefore Q1 is (50.5 – 80.5)
= L+ h/f(iN/4-c)
= 50.5+ 30/12(1(40)/4-6)
= 50.5 + 10
= 60.5 or 61
Q3 = 3(N)th value
= (3(40))/4 th value
= 30th value
30th value is in the interval 80.5 – 110.5
Therefore Q3 is (80.5 – 110.5)
= L+ h/f(iN/4-c)
= 80.5+ 30/14(3(40)/4-18)
= 80.5 + 25.71
= 106.21
The Quartile Deviation = (Q3 – Q1)/2
= (106.2-60.5)/2
= 22.85 or 23
QUESTION 2
2.1 Mean (μ)
= X/N
= 10833/12
= 902.75
Standard Deviation (σ) = √(〖(X-M)〗^2/(N-1))
=√(143420/(12-1))
= √13038.21
= 114.185
= P(X < 800) = P(X -)/ < (800 - 902.75)/114.185 = P (Z< -0.899855) = .1841 1 billion = 1000 million μ = 902.75 σ = 114.185 standardise x to z = (x - μ) / σ P(x > 1000) = P (z > (1000-902.75) / 114.185)
= P (z > 0.8517) = .1977
We know from the standard normal that
P (Z < z) = 0.95 = P(Z > z) = 0.05
when z = 1.644854
x = μ + z * σ
x = 902.75 + 1.644854 * 114.185
x = 1090.568 or 1091
QUESTION 3
H0: There is no evidence of a change in the market shares of car rental firms
H1: There is evidence of a change in the market shares of car rental firms
 = 0.05
Reject H0 if the calculated X2 > 15.507
df = (k-1)(r-1) = (5-1)(3-1) = 8
CONTINGENCY TABLE:
CONTINGENCY TABLE
x^2=  ((f0 – fe)2)/fe
28.91
0.041
24.78
0.002
16.84
0.593
17.47
0.691
0.140
30.13
0.042
25.83
0.309
17.55
0.012
18.21
0.176
0.404
31.96
0.000
27.39
0.248
18.61
0.701
19.31
0.147
0.068
3.575