Statistics Assignment-2 (Weeks: 5-7) | Desklib

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This is the solution to Statistics Assignment-2 (Weeks: 5-7) for the 1st Semester, 1439-1440 (2018-2019) at CSTS-SEU-KSA. The assignment covers probability distribution, variance, binomial distribution, normal distribution, confidence interval for population proportion and confidence interval for population mean. The solution is provided with step-by-step calculations and formulas used.

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CSTS-SEU-KSA
___________________________________________________________________
Statistics (STAT-101)
Assignment-2 (Weeks: 5-7)
1st Semester, 1439-1440 (2018-2019)
Due date: 20/10/2018 (Time: 10:30 PM)
Student’s Name
Student’s ID
Section/CRN
Location
Marking Scheme
Question Score Obtained Score
Q-1 3
Q-2 3
Q-3 3
Q-4 3
Q-5 3
Q-6 3
Total 18
Note: You are required to fill your full name, ID and CRN.
Solution to the Questions

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CSTS-SEU-KSA
Question One
a. Probability distribution table for the random variable
Solution
Type of a Coin Frequency
f
Probability
p= f
f
$ 1 2 0.22
$ 5 3 0.33
$ 10 1 0.11
$ 20 3 0.33
Total() 9 1.00
b. Variance.
According to Hassett & Stewart (2006), the variance is given by the formula
Var ( x )=E ( x2 )¿ ¿
¿ x2 ¿ px ( xpx )
2
The table below summarizes the result of the computation
Type of a Coin Frequency
f
x=coin $f Probability
px
xpx x2px
$ 1 2 2 0.22 0.444 0.889
$ 5 3 15 0.33 5.000 75
$ 10 1 10 0.11 1.111 11.111
$ 20 3 60 0.33 20.000 1200
Total() 9 1.00 26.556 1287
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Therefore,
¿ x2 ¿ px ( xpx )
2
=1287 ( 26.556 ) 2
¿ 1287705.198=581.802 .
Therefore, the variance is $ 581.80
Question Two
Solution
The probability of interest isP rob ( x 3 )=1P ( x <3 )
The distribution of the 5 students adopts binomial distribution with probability of
success ( p ) of an event been 0.13 and that of failed ( q ) event is 0.87. The number of
independent trials ( n ) is 5 and the number of success 3or more ( r )
According to Hassett & Stewart (2006), the probability of success in a binomial
distribution is given by
P rob ( x=r ) = [ n
r ] ( p ) r ( q ) nr
Therefore,
P rob ( x 3 ) =1P rob ( x<3 )
¿ 1 {P rob ( x=0 ) + P rob ( x=1 )+ P ( x=2 ) }
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P rob ( x=0 )= [5
0 ] ( 0.13 )0 ( 0.87 )50= ( 0.87 )5=0.498
P rob ( x=1 ) =
[ 5
1 ] ( 0.13 )1 ( 0.87 ) 51=50.13( 0.87 ) 4=0.372
P rob ( x=2 ) =[ 5
2 ] ( 0.13 ) 2 ( 0.87 )5 2 =50.0169( 0.87 )3 =0.111
Thus,
1 { P rob ( x=0 ) + P rob ( x=1 ) + Prob ( x=2 ) } =1 {0.498+ 0.372+ 0.111 }
¿ 0.01 8
Hence, P rob ( x 3 )=0.018
Question Three
Solution
The answer will be obtained from the normal distribution tables
a. P rob( z< 2.37)
P rob ( z <2.37 )=0.991 1
b. P rob(z>1.18)
P rob ( z >1.18 )=1P rob ( z <1.18 )
¿ 10.119=0.881

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CSTS-SEU-KSA
Thus, P rob ( z >1.18 )=0.881
c. P rob(1.18< z <2.37)
P rob (1.18< z<2.37 )=P rob ¿
¿ 0.991 10.119=0.8721
Hence, P rob (1.18< z<2.37 )=0.8721
Question Four
Solution
The probability of interest is P rob ( x 7.995 )=1P rob ( x <7.995 )
From the data given, the population mean ( μ ) =8.025 and
Population standard devition ( σ )=0.125
The probability will be computed by using z-scores
In this case, he z-score will be given by the formula
z= xμ
σ
n
, where n=sample ¿ 36
1P rob ( x<7.995 ) =1P rob
{z < 7.9958.025
0.125
36 }
¿ 1P rob {z < 0.03
0.02083 }=1P rob { z 1.44 }
¿ 10.0749=0.925 1
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Therefore, P rob ( x 7.995 )=0.9251
Question Five
Solution
According to Goos & Meintrup (2016), the confidence interval for the population proportion
( ^pu ) is given by
^pu= ^p ± z
^p ( 1 ^p )
n ,
where n=sample=150 , ^p=sample proportion= 60
150 =0.4 0
z=multiplier ¿ z table=1.96 at 95 %
^pu=0.4 ±1.96
0.4 ( 10.4 )
150 =0.4 ± 1.96 0.24
150
¿ 0.4 ± 1.960.04=0.4 ± 0.078 4
Therefore the confidence interval for the population proportion cholesterol level of 200 is
(0.3216, 0.4784).
Question Six
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Solution
According to Goos & Meintrup (2016), the 95% confidence interval of the population mean ( μ )
is constructed by the formula;
μ= x ± zσ
n , where x=smaple mean=7.10 , n=sample ¿200
z=multiplier ¿ z table=1.96 at 95 %σ= population standard deviation
¿ 7.10 ± 1.965
200 =7.10 ±1.960.353 6
¿ 7.10 ± 0.693
Therefore, the 95% confidence interval for μ is ( 6.407 , 7.793 )

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Reference
Goos, P., & Meintrup, D. (2016). Statistics with JMP: Hypothesis Tests, ANOVA and
Regression. John Wiley & Sons.
Hassett, M. J., & Stewart, D. (2006). Probability for risk management. Actex Publications.
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