The following circle graph shows how Brad spends his time (excluding sleeping) each week outside of school. Which of the followi
ng statements is true about how Brad spends his time?
A.Brad spends more time working than he does on homework.
B.Brad spends about the same amount of time watching videos and reading combined as he does playing the piano and soccer combined.
C.Brad spends about the same amount of time doing homework as he does watching videos.
D.Brad spends about the same amount of time playing piano as he does playing football and soccer combined.
A.Brad spends more time working than he does on homework.
B.Brad spends about the same amount of time watching videos and reading combined as he does playing the piano and soccer combined.
C.Brad spends about the same amount of time doing homework as he does watching videos.
D.Brad spends about the same amount of time playing piano as he does playing football and soccer combined.
1 answer:
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Answer:
y 0 1 2 3
P(Y=y) 0.0676 0.3549 0.3875 0.19
Step-by-step explanation:
P(Wed) = 0.26
P(Thurs) = 0.39
P(Fri) = 0.25
P(Sat) = 0.10
Y = No. of days beyond Wednesday it takes for both magazines to arrive i.e. 0,1,2,3
Y=0 means the magazines will arrive on Wednesday
Y=1 means the magazines will arrive till Thursday
Y=2 means the magazines will arrive till Friday
Y=3 means the magazines will arrive till Saturday
The possible combinations for Y are
Y(W,W) Y(W,T) Y(W,F) Y(W,S)
Y(T,W) Y(T,T) Y(T,F) Y(T,S)
Y(F,W) Y(F,T) Y(F,F) Y(F,S)
Y(S,W) Y(S,T) Y(S,F) Y(S,S)
So, we can classify these possible outcomes as Y=0,1,2,3.
Y(0) = Y(W,W) (both magazines take 0 days to arrive beyond Wednesday)
Y(1) = Y(W,T), Y(T,T), Y(T,W) (both magazines take 1 day to arrive beyond Wednesday)
Y(2) = Y(W,F), Y(T,F), Y(F,F) Y(F,W) Y(F,T) (both magazines arrive till Friday)
Y(3) = Y(W,S), Y(T,S), Y(F,S), Y(S,W), Y(S,T), Y(S,F), Y(S,S) (both magazines arrive till Saturday)
To calculate the PMF, we need to calculate the probability for each of the points in Y(0,1,2,3).
Y(0) = Y(W,W)
= 0.26 x 0.26
Y(0) = 0.0676
Y(1) = Y(W,T) + Y(T,T) + Y(T,W)
= (0.26 x 0.39) + (0.39 x 0.39) + (0.39 x 0.26)
= 0.1014 + 0.1521 + 0.1014
Y(1) = 0.3549
Y(2) = Y(W,F) + Y(T,F) + Y(F,F) + Y(F,W) + Y(F,T)
=(0.26 x 0.25) + (0.39 x 0.25) + (0.25 x 0.25) + (0.25 x 0.26) + (0.25 x 0.39)
= 0.065 + 0.0975 + 0.0625 + 0.065 + 0.0975
Y(2) = 0.3875
Y(3) = Y(W,S) + Y(T,S) + Y(F,S) + Y(S,W) + Y(S,T) + Y(S,F) + Y(S,S)
= (0.26 x 0.10) + (0.39 x 0.10) + (0.25 x 0.10) + (0.10 x 0.26) + (0.10 x 0.39) + (0.10 x 0.25) + (0.10 x 0.10)
= 0.026 + 0.039 + 0.025 + 0.026 + 0.039 + 0.025 + 0.010
Y(3) = 0.19
y 0 1 2 3
P(Y=y) 0.0676 0.3549 0.3875 0.19
The PMF plot is attached as a photo here.
