Math DB 3

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School

Fayetteville State University *

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210

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Statistics

Date

Apr 3, 2024

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1. Choose a value for p between 0.20 and 0.80. It should have at least two decimal places. When you create the thread on Blackboard, put your p = your value in the title. I must be different from the other student's p. P=.444 2. Find the following binomial probabilities. Show all your work. Round to 4 decimal places. a. Find the probability that exactly 6 of 25 are successes. Binompdf(25,.444,6)= 0.0295 b. Find the probability that at least 10 of 25 are successes. 1-Binomcdf(25,.444,9)= 0.7383 c. Find the probability that at most 5 of 25 are successes. Binomcdf(25,.444,5)= 0.0100 d. Find the probability that there are between 3 and 7 successes, inclusive. Binomcdf(25,.444,7)- Binomcdf(25,.444,3)= 0.0716 TI 84, 2 nd , VARS (Distr) Find the probability that exactly 6 of 25 are successes. a. Find the probability that at least 10 of 25 are successes. b. Find the probability that at most 5 of 25 are successes. c. Find the probability that there are between 3 and 7 successes, inclusive Mathew Post 2 a. binompdf(25, .33,6) = .1134 b. 1-binomcdf(25, .33 ,10) = .1686 c. binocmdf(25, .33,5) = .1187 d. binomcdf(25, .33 ,7) - binomcdf(25, .33,3) = .3676 I checked your answer using my calculator my answers are above. Normal Approximation: N=25, P=.33 Q= 1-.33=.67 Finding NP and NQ to make sure they are greater than 5. NP= (25)(.33)=8.25 NQ=(25)(.67)=16.75

Mean= 8.25 Standard deviation is the square root of (25)(.67)(.33)=1.1055 1: The probability of 6 and 25 P(n-0.5< x >n+0.5)= 5.5, 6.5 Z1 = 5.5-8.25/1.1055= .2261 Z2= 6.5-8.25/1.1055 =-1.5830 Normalcdf(-9999,.2261,0,1) - Normalcdf(-9999,-1.5830,0,1)= .5327 2: The probability that at last 10 of 25 are successful We have to use 9.5 Z1= 9.5-8.25/1.1055= 1.1307 Normalcdf(-9999,18.0914,0,1)= .8709 3: To find the probability of at most 5 are successful We have to use 5.5. Z1= 5.5-8.25/1.1055= -2.4876 Normalcdf(21.7096,9999,0,1)= .9936 4: To find the probability of success between 3 and 7 We have to use 2.5 and 7.5 Z1=2.5-8.25/1.1055= -5.2013 Z2=6.5-8.25/1.1055= -1.5830 Normalcdf(-9999,24.4233,0,1) -Normalcdf(-9999,20.8051,0,1)= -.0567 Luis Post 3 In post one Bionompdf(25,.26,6)= .1793 1-Bionomcdf(25,.26,9)= .0893 Bionompdf(25,.26,5)= .3356 Bionomcdf(25,.26,7)- Bionomcdf(25,.24,3)= .6069 In post two Normalcdf(-9999,.5.5,0,1) -Normalcdf(-9999.6.5,0,1) =0.1760 Normalcdf(9.5,9999,0,1)= 0.0854 Normalcdf(-9999,4.5,0,1)= 0.1806 Normalcdf(-9999,6.5,0,1) - Normalcdf(-9999,2.5,0,1)= 0.6421 In conclusion of all this information of post 1 and post 2, the normal approximation is very close together. The binomial distribution is discreate and is not possible to find the exact date between any values. The normal distribution is continuous with a bell shape and any

two data point could alien with another. In this data collection the normal distribution and the binomial distribution are closely related based on the similarity of the answers.

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