Question

A certain bowling league has weekly matches where two teams compete against each other. Each team has four players whose individual scores are combined to form their team's overall score that week. Based on data from previous matches, the table below shows summary statistics for the weekly scores of two teams.
Team Mean Standard deviation
Bowling Buds \mu_B=610μ
B
​
=610mu, start subscript, B, end subscript, equals, 610 \sigma_B=60σ
B
​
=60sigma, start subscript, B, end subscript, equals, 60
Alleycats \mu_A=640μ
A
​
=640mu, start subscript, A, end subscript, equals, 640 \sigma_A=80σ
A
​
=80sigma, start subscript, A, end subscript, equals, 80
Let A represent the Alleycats' score on a randomly chosen week, B represent the Bowling Buds' score on a randomly chosen week, and D represent the difference between their scores (D=A-B)left parenthesis, D, equals, A, minus, B, right parenthesis.
Find the mean of D.

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Answer 1

Answer: 30 Pins

Step-by-step explanation:

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Answer 2

Subtracting their means, it is found that the mean of D is of 30.

• When two variables are subtracted, the mean is the subtraction of the means, while the standard deviation is the square root of the sum of the variances.

In this problem, the means of the variables are:

$\mu_B = 610, \mu_A = 640$

Since D = A - B, the mean of D is:

$\mu_D = \mu_A - \mu_B = 640 - 610 = 30$

The mean of D is of 30.

A similar problem is given at https://brainly.com/question/16371181