Question

Two friends, Andy and Bob, participate in a game of bowling every week. From past experiences, it is known that both friends’ scores are approximately normally distributed, where Andy has a mean score of 150 with a standard deviation of 30, and Bob has a mean score of 165 with a standard deviation of 15. Assuming that their scores are independent, which of the following values is closest to the probability that Andy will have a greater score than Bob in a single game?

Answer 1

Answer: 0.3264

Step-by-step explanation:

Let Andy's variable = a

Bob's variables = b

Andy mean score Ma = 150

Andy standard deviation SDa = 30

Bob mean score Mb = 165

Bob standard deviation = 15

Therefore, the probability of Andy having greater score than Bob

= P( a-b > 0)

The mean X(a-b) = Ma - Mb

X(a-b)= 150 - 165 = -15

Standard deviation = sqrt Variance

Variance = 30² + 15² = 1125

SD = sqrt (1125) = 33.54

Therefore, the

P(a-b > 0) = 1 - P(a-b <0)

P(a-b<0) = P(z, 0-(-15)/33.54 )

P(a-b <0) = P (z, 0.45) = 0.6736

Since

P(a-b > 0) = 1 - P(a-b <0)

P(a-b> 0) = 1 - 0.6736 = 0.3264

Therefore, Andy has the probability of 0.3264 of scoring greater Bob

Answer 2

Andy has the probability of 0.3264 scoring greater than Bob.

Let Andy's variable = a

Bob's variables = b

Andy mean score Ma = 150

Andy standard deviation SDa = 30

Bob mean score Mb = 165

Bob standard deviation = 15

Therefore, the probability of Andy having greater score than Bob

= P( a-b > 0)

The mean X(a-b) = Ma - Mb

X(a-b)= 150 - 165 = -15

Standard deviation = sqrt Variance

Variance = 30² + 15² = 1125

SD = sqrt (1125) = 33.54

Therefore, P(a-b > 0) = 1 - P(a-b <0)

P(a-b<0) = P(z, 0-(-15)/33.54 )

P(a-b <0) = P (z, 0.45) = 0.6736

Since, P(a-b > 0) = 1 - P(a-b <0)

P(a-b> 0) = 1 - 0.6736 = 0.3264

Therefore, Andy has the probability of 0.3264 of scoring greater than Bob.

For more information:

https://brainly.com/question/14116780