Asked by NIck
true/false test with 16 questions all answers are guessed what is the mean and std deviation can u tell me how to get the P(x) value
Answers
Answered by
Damon
This is a binomial distribution with p = 1/2 and n = 16
p
mean = n p = 8
sigma^2 = n p(1-p) = 16*(1/2)(1/2) = 4
so sigma = 2
to get probability of k trues (or false since p = 1/2 and therefore 1-p also = 1/2
P(k) = C(n/k)* p^k *(1-p)^(n-k)
C(n/k) is "binomial coef. Look it up in table or use Pascal's triangle or use
C(n/k) = n!/[k!(n-k)!]
In our case (1/2 is much easier than any other)
p^k * (1-p)^(n-k) =(1/2)^k * (1/2)^(16-k) = (1/2)^16 = 1/65536
our C(16/k) values are
1 16 120 560 1820 4368 8008 11440 12870 11440 8008 4368 1820 560 120 16 1
note that the first value, 1, is for n = 16, k = 0
so for example the probability of 6 true is 8008/65536 = .122
P(4) = 1820
p
mean = n p = 8
sigma^2 = n p(1-p) = 16*(1/2)(1/2) = 4
so sigma = 2
to get probability of k trues (or false since p = 1/2 and therefore 1-p also = 1/2
P(k) = C(n/k)* p^k *(1-p)^(n-k)
C(n/k) is "binomial coef. Look it up in table or use Pascal's triangle or use
C(n/k) = n!/[k!(n-k)!]
In our case (1/2 is much easier than any other)
p^k * (1-p)^(n-k) =(1/2)^k * (1/2)^(16-k) = (1/2)^16 = 1/65536
our C(16/k) values are
1 16 120 560 1820 4368 8008 11440 12870 11440 8008 4368 1820 560 120 16 1
note that the first value, 1, is for n = 16, k = 0
so for example the probability of 6 true is 8008/65536 = .122
P(4) = 1820
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