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Running Head: STATISTICS
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Running Head: STATISTICS
Question one:
0.95 chance of winning nothing
0.02 chance of winning $1.00
0.02 chance of winning $5.00
0.01 chance of winning $20.00
Required: Average winning per game. (0.02*0.02*0.01)/3=0.000001333
Question 2:
The Binomial Probability density where p=0.3 and n=3, then q=1-0.3=0.7
3C0(0.3)0*(0.7)3+3C1(0.3)1(0.7)2+3C2(0.3)2(0.7)1+3C3(0.3)3(0.7)0=0.343+0.441+0.189+0.027=1
Question 3:
The Probability of drawing a red card or an ace. Let B be the event that the card drawn is an
ace and Q be the event that the card drawn is red.
P(B and Q)=1/26
P(B)=1/26
P(Q)=1/2
P(B or Q)=P(B)+P(Q)-P(B and Q)
P(B or Q)=(1/13)+(1/2)-(1/26)=14/26=7/13
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Running Head: STATISTICS
Question 4:
Is a normal random variable discrete or continuous?
A normal random variable is continuous in nature. This is because it follows the normal
distribution which is continuous and takes values which are infinite and uncountable.
Question 5:
What is his expected score?
Possible outcomes are 5: A,B,C, D, E
µ=∑xf(x), where x=50 and f(x)=1/5 hence µ=50*1/5=10
Question 6:
The probability that more than 180 villagers die from the disease?
µ=160=(0.49*326)
δ=√npq=√326*0.51*0.49=9.025929315
x=180
Z=(180-160)/9.025929315=2.22
Corresponding area=0.4868
P(X>180)=(0.5-0.4868)=0.0132
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Running Head: STATISTICS
Question 7:
Probability that a random patient will infect 2 or more patients.
µ=∑xf(x) where f(x) is the probability of infection and x=2 or more
1.75=2f(x)=0.875
Hence P(x≥2)=0.875
Question 8:
Percentage of people who contract the disease will show symptoms after 21 days?
µ=10 days
δ=4 days
x=21
Z=(21-10)/4=2.75
Area under z is 0.4970
P(x>21)=(0.5-0.4970)=0.003
0.003*100=0.3% will sow symptoms after 21 days.
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Running Head: STATISTICS
Question 9: 95% confidence interval for the mortality rate
Proportion of success(4493/8999)=0.499277697
n =8999, ZC=1.96
95% Confidence interval= proportion of success±1.96*(√pg/n)
0.499277697±1.96*(√0.499277697*0.500722302/8999)
0.499277697±0.01033067
0.488947027<p<0.509608367
Question 10:
Probability that first visit to free parking takes place after the second trip
Application of multiplication rule
P(x)=0.02*0.02=0.0004
Question 11: The data is run on Statdisk and the results obtained.
Sample mean=(31+42+38+43+39+40)/6=38.83333333
Sample
variance=((38.83-31)2+(38.83-42)2+(38.83-38)2+(38.83-43)2+(38.83-39)2+(38.83-
40)2)/5)=18.16668
Sample standard deviation=√18.16668=4.262238848
Alfa=0.01
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Running Head: STATISTICS
99% Confidence interval is given by: Sample mean±(talfa/2*s/√n-1) where n-6
38.383333333±(4.032*4.262237/√5), which gives use 38.383333333±7.685517505
Hence the 99% Confidence interval is between: 30.69781583<µ<46.06885084
Question 12:
n =49.
µ=8
δ=5, the sample mean has a normal distribution. This is because the sample size n>30 and the
population standard deviations are know.
Question 13:
Percentage of normal data from population that lies between 0 and 7.
Z1=(0-3)/5=-0.7 Corresponding area=0.2257
Z2=(7-3)/5=0.8, Corresponding area=0.2881
Percentage
of
normal
data
from
population
that
lies
between
0
and
7=(0.2881+0.2257)=0.5138*100=51.38%
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