Question

Angelita and Cynthia like to play a video game against each other that is scored by round. The goal is to score as many points as possible in each round. Here are summary statistics how many points they eacth score per round:

Mean S tandard deviation
Angelita μA-900 σA = 96
Cynthia μC 800 σC = 72

Both distributions are approximately normal. Suppose we choose a round at random and calculate the difference between their scores. We can assume that their scores each round are independent

a. Find the probability that Angelita's score is higher than Cynthia's. You may round your answer to two decimal places.

Answer 1

To determine the probability that Angelita's score is higher than Cynthia's, the difference in their scores is modeled as a normal distribution. The mean difference in their scores is 100 and the standard deviation is 120. The z-score for the difference being greater than 0 is -0.8333, leading to an approximate probability of Angelita scoring higher than Cynthia as 0.80 or 80%.

Step-by-step explanation:

To find the probability that Angelita's score is higher than Cynthia's, we need to consider the distribution of the difference in their scores. If we denote the random variable D = Angelita's score - Cynthia's score, then D is normally distributed since we are given that both Angelita's and Cynthia's individual scores are approximately normally distributed with known means and standard deviations, and we are assuming that their scores are independent.

The mean of D, denoted as μD, is the difference between the means of Angelita's and Cynthia's scores, so μD = μA - μC = 900 - 800 = 100.

The standard deviation of D, denoted as σD, can be found using the formula for the standard deviation of the difference of two independent random variables, σD = √(σA2 + σC2) = √(962 + 722) = √(9216 + 5184) = √(14400) = 120.

Now we want to find the probability that D is greater than 0, which corresponds to Angelita scoring higher than Cynthia. The z-score for D = 0 is calculated as Z = (0 - μD) / σD = (0 - 100) / 120 = -0.8333.

Using the standard normal distribution, we can then find the probability that Z > -0.8333, which is the same as the probability that Angelita scores higher than Cynthia. Consulting the standard normal cumulative distribution tables or using a statistical calculator, we find that P(Z > -0.8333) ≈ 0.80 (approximated to two decimal places).

Therefore, the probability that Angelita's score is higher than Cynthia's in a randomly chosen round is approximately 0.80 or 80%.