ng18-Ma × a Secure | https://webworks.frostburg.edu/webwork2/Spring18-Math380/03

Question

ng18-Ma × a Secure | https://webworks.frostburg.edu/webwork2/Spring18-Math380/03.05 Preview/1/ 03.05 Preview: Problem 1 Prev Next Entered 0.200126962 0.3 0.15 0.2054 Prev 0.3 0.15 0.2054 0.55 At least one of the answers above is NOT correct (1 pt) Suppose the number of cars in a household has a binomial distribution with parameters n-12, and p 65 % Find the probability of a household having: (a) 3 or 9 cars 0.200126962 (b) 7 or fewer cars 0.3 (c) 4 or more cars 0.15 (d) fewer than 9 cars 0.2054 (e) more than 7 cars 0.55 Note: You can earn partial credit on this probiem search

Explanation / Answer

As per binomial distribution,
P(X=r) = nCr * p^r * (1-p)^(n-r)

a)
0.200126962
Excel formula: =BINOM.DIST(3,12,0.65,FALSE)+BINOM.DIST(9,12,0.65,FALSE)

b)
P(X<=7) = 0.41665495
Excel formula: =BINOM.DIST(7,12,0.65,TRUE)

c)
P(X >= 4) = 1 - P(X<=3)
P(X<=3) = 0.0056097
P(X >= 4) = 1 - 0.0056097 = 0.99439024
Excel formula: =1-BINOM.DIST(3,12,0.65,TRUE)

d)
P(X < 9) = 0.65334730
Excel formula: =BINOM.DIST(8,12,0.65,TRUE)

e)
P(X > 7) = 0.58334505
Excel formula: =1-BINOM.DIST(7,12,0.65,TRUE)

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