Find the Probablilites for a sample of 20 teenagers of 70% of them had compact disk players by the age of 16?

Question

at least 14 had cd players. Exactly 9 had cd players. more than 17 had cd players. Which event a,b,c is most likely to occur?

Answer 1

(a) P(at least 14) = P(14) + P(15) + P(16) + ... + P(20)
p = .7 q = 1-p = 1-.7 = .3
n = # of total teenagers = 20
k = # of teenagers that had disk players by the age of 16
n-k = # of teenagers that didn't have disk players by the age of 16

Equation
n!/k!(n-k)! * p^k * q^(n-k)
P(at least 14) = 20!/(14!6!) *.7^14 * .3^6 + 20!/(15!5!) *.7^15 * .3^5 + 20!/(16!4!) *.7^16 * .3^4 + 20!/(17!3!) *.7^17 * .3^3 + 20!/(18!2!) *.7^18 * .3^2 + 20!/(19!1!) *.7^19 * .3 + 20!/(20!) *.7^20 = .608

(b) P(exactly 9) = P(9) Using equation above
20!/(9!11!) * .7^9 * .3^11 = .012

(c) P(more than 17) = P(18) + P(19) + P(20) Again using equation from (a)
20!/(18!2!) *.7^18 * .3^2 + 20!/(19!1!) *.7^19 * .3 + 20!/(20!) *.7^20 = .035

Based on the answers, event (a) is most likely to occur.

Answer 2

70% of 20 is 14. There's no "at least" about it. If 70% of 20 had CD players, then 14 had them.

Answer 3

At least 14 means:

Either 14 or 15 or 16 or 17 or 18 or 19 or 20

14 had cd players + 15 had cd players + etc.

20C14 (0.7^14 )(0.3^6) + 20C15 (0.7^15)(0.3^5) etc....

Exactly 9

20C9 (0.7^9)(0.3^11)

More than 17 means

Either 18 or 19 or 20

18 had cd
20C18 (0.7^18)(0.3^2) + etc.


Without computing, the first choice seems to be most likely. If you aren't sure, compute the above. It is tedious but it is most reliable. (Binomial distribution)

Answer 4

20 x 70% = 14
the answer is neither a, nor b, nor c...
it should be "exactly 14 had cd players".

Answer 5

honey,,almost everyone has ipods at that age. not cd players. its not the 90s

Answer 6

derr......you need to just relax a bit. have a beer or three...cause who could possibly care? im guessing youre a uni student......

Answer 7

A

Answer 8

i think it is b
what a random question :-)
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