Graded Due at the start of class on Monday June 4m ote: Tutoriol assistance is a

Question

Graded Due at the start of class on Monday June 4m ote: Tutoriol assistance is alowed Problem #4 (The Last One O) (See 01) or Tuesday lune 5" plete the assignment according to the ent. Remember an answer withements in the "Expectations for All Graded Work work or explanation will not receive full credit 1. Tainted Currency Based on American Chemical Society randomly selected dollar bill is tainted research, there is a o.90 probability that in the United States a randomly selected. Report your answers using with traces of cocaine. Assume that ten dollar bills are corre ct probability notation. Hint: Is this a binomial random variable? a. Find the probability that all 10 bills have traces of cocaine. b. Find the probability that exactly 5 bills have traces of cocaine c. Find the probability that the majority (i.e. more than half) of the bills have traces of cocaine. Purchasing a Home in Upstate New York A quantitatively savvy, young couple is interested in 2. the area. They want to predict the selling price of homes (in thousands of dollars) based on the size of the home (in square feet). The regression equation is Price -86.097+0.248 Size t Stat P-value Regression Statistics CoeficientsStandard Multiple R intercept 86.097322 82.2443144 -1.0468483 0.30559862 0.745865819 Size 0.24792500 0.04519506 5.48566590 1.22193E-05 R Square055631582 Adjusted R Square 0.537828979 162.8663093 26 Standard Error 1200 Observations 1000 800 600 400 200 0 500 1000 1500 2000 2500 3000 3500 Size in Square Feet -over

Explanation / Answer

Answer 1

we have probability p = 0.90, total sample(n) = 10

(a)

we have to find the probability that all 10 bills have traces of cocaine

P(x=10) = 1 - P(x=0)

so, we need to find the probability that exactly 0 bills have traces of cocain, then we will get the required probability that all 10 bills have traces of cocaine

using binomial distribution formula, we can write

P(x=r) = C(n,r)*(p^r)*(1-p)^(n-r)

setting n = 10, p = 0.90 and r = 0, we get

P(x=0) = C(10,0)*(0.9^0)*(1-0.9)^(10-0) = 10!/[(10-0)!*0!]*(0.9^0)*(0.10)^(10) = 1*10^10 or less than 0.000001

so, the probability that all 10 have traces of cocaine = 1 - 0.000001 = 0.999999

(b)

using binomial distribution formula, we can write

P(x=r) = C(n,r)*(p^r)*(1-p)^(n-r)

setting n = 10, p = 0.90 and r = 5 because we want to the probability of exactly 5, we get

P(x=5) = C(10,5)*(0.9^5)*(1-0.9)^(10-5) = 10!/[(10-5)!*5!]*(0.9^5)*(0.10)^(5) = 0.001488 or 0.0015

so, probability of exactly 5 is 0.0015

(c)

Probability that the majority of the bills have traces of cocain means we have to find the probability of 5 or more bills to have traces of cocaine

so, we need to calculate the probability(5 or more) = P(x=5)+P(x=6)+P(x=7)+P(x=8)+P(x=9)+P(x=10)

it can be solved as

P(x=5) = C(10,5)*(0.9^5)*(1-0.9)^(10-5) = 10!/[(10-5)!*5!]*(0.9^5)*(0.10)^(5) = 0.001488 or 0.0015

P(x=6) = C(10,6)*(0.9^6)*(1-0.9)^(10-6) = 10!/[(10-6)!*6!]*(0.9^6)*(0.10)^(4) = 0.0112

P(x=7) = C(10,7)*(0.9^7)*(1-0.9)^(10-7) = 10!/[(10-7)!*7!]*(0.9^7)*(0.10)^(3) = 0.0574

P(x=8) = C(10,8)*(0.9^8)*(1-0.9)^(10-8) = 10!/[(10-8)!*8!]*(0.9^8)*(0.10)^(2) = 0.1937

P(x=9) = C(10,9)*(0.9^9)*(1-0.9)^(10-9) = 10!/[(10-9)!*9!]*(0.9^9)*(0.10)^(1) = 0.3874

and

P(x=10) = C(10,10)*(0.9^10)*(1-0.9)^(10-10) = 10!/[(10-10)!*10!]*(0.9^10)*(0.10)^(0) = 0.3487

adding all these probabilities, we get P(5 or more) = 0.9999

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