Answer:
A) from the line of best fit, the approximately y-intercept is (0,1.8). This means without any practice, 1h.8 games are won.
B) slope: (5.6-1.8)/(2-0) = 1.9
y = 1.9x + 1.8
(Line of best fit)
x = 13,
y = 1.9(13) + 1.8 = 26.5
Predicted no. of games won after 13 months of practice is 26.5
0 lightbulbs would be expected to burn out in 1800 hours
The expected completion time is μ = 40 weeks.
The random variable, X = 38 weeks (the probable time)
If the standard deviation is σ, then the z-score is z = (x - μ)/σ.
Let us test the given standard deviations.
When σ=1,
z = (38-40)/1 = -2
From standard tables,
P(x<=38) = P(z<=-2) = 0.0228 =2.3%
When σ=2,
z = (38*40)/2 = -1
P(x<=38) = P(z<=-1) = 0.1587 = 15.9%
When σ=4,
z = (38-40)/4 = -0.5
P(x<=38) = P(z<=-0.5) = 0.3085 = 30.9%
Answers:
z=1 => 2.3%
z=2 => 16% (approx)
z=4 => 31% (approx)
Y = 16 because 4x3 = 12 + 4 = 16
