Considering the data the number 1, find the mean variance standard deviation of

Question

Considering the data the number 1, find the mean variance standard deviation of the data, and the z-scores of each of the die outcomes (1, 2, ..., 6) based on the mean and standard deviation. (Round your values to three decimal places) Use the data in the number 1, find the followings. a. P(1), P(2), P(3), P(4), P(5), P(6) b. P(Even), P(odd), P (multiples of 3), P(Prime number) c. P(Even Intersection multiples of 3), P(odd Union Prime number) d. P(oddPrime number) e. Are the events 'Even' and 'Prime number independent? Why?

Explanation / Answer

P(x) = no of successful outcomes/total no of outcomes

(a)

p(1) = 17/120 = 0.14

p(2) =16/120 = 0.13

p(3) = 20/120 = 0.16

p(4) = 19/120 = 0.15

p(5) = 23/120 = 0.19

p(6) = 25/120 = 0.208

(b)

p(even) = p(2) + p(4) + p(6)

= 0.13 + 0.15 + 0.208 = 0.488

p(odd) = p(1) + p(3) + p(5)

= 0.14 +0.16 +0.19 = 0.49

p(multiples of 3) = p(3) + p(6)

= 0.16 + 0.208 = 0.368

p(prime no) = p(2)+p(3)+p(5)

= 0.13 + 0.16 +0.19 = 0.48

(c)

p(even multiples of 3) = p(x is even) and p(x is multiple of 3) = p(6) = 0.208

p(odd prime no) = p(x is odd) or p(x is prime) = p(1)+p(2)+p(3)+p(5) = 0.62

(d)

p(a/b) = p(a b)/p(b)

p(odd/prime no) = p(odd prime)/p(prime)

= p(3)+p(5)/p(2)+p(3)+p(5)

=0.16+0.19 / 0.48

=0.35/0.48 = 0.73

(e)

p(even prime) = p(even and prime) = p(2) = 0.13

p(even) * p(prime) = 0.48 * 0.48 = 0.23

since p(even prime) is not equal to p(even) * p(prime), they are not independent.

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