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Running Head: STATISTICS Student Name: Institution: Professor: Course Title: Date of Submission: 1 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 2 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 3 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. 4 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 5 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% 6

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