It's hard to tell where one set ends and the next starts. I think it's
A. 25, 36, 44, 51, 62, 77
B. 3, 3, 3, 7, 9, 9, 10, 14
C. 8, 17, 18, 20, 20, 21, 23, 26, 31, 39
D. 63, 65, 66, 69, 71, 78, 80, 81, 82, 82 Let's go through them.
A. 25, 36, 44, 51, 62, 77
That looks OK, standard deviation around 20, mean around 50, points with 2 standard deviations of the mean.
B. 3, 3, 3, 7, 9, 9, 10, 14
Average around 7, sigma around 4, within 2 sigma, seems ok.
C. 8, 17, 18, 20, 20, 21, 23, 26, 31, 39
Average around 20, sigma around 8, that 39 is hanging out there past two sigma. Let's reserve judgement and compare to the next one.
D. 63, 65, 66, 69, 71, 78, 80, 81, 82, 82
Average around 74, sigma 8, seems very tight.
I guess we conclude C has the outlier 39. That one doesn't seem like much of an outlier to me; I was looking for a lone point hanging out at five or six sigma.
The mean of the bell shaped fluorescent light bulb μ = 61
The standard deviation σ = 11
The objective of this question is to determine the approximate percentage of light bulb replacement requests numbering between 61 and 94 i.e P(61≤ X≤94)
Using the empirical (68-95-99.7)rule ;
At 68% , the data lies between μ - σ and μ + σ
i.e
61 - 11 and 61 + 11
50 and 72
At 95%, the data lies between μ - 2σ and μ + 2σ
i.e
61 - 2(11) and 61 + 2(11)
61 - 22 and 61 +22
39 and 83
At 99.7%, the data lies between μ - 3σ and μ + 3σ
i.e
61 - 3(11) and 61 + 3(11)
61 - 33 and 61 + 33
28 and 94
the probability equivalent to 94 is when P(28≤ X≤94) =99.7%
This implies that ,
P(28≤ X≤94) + P(61≤ X≤94) = 99.7%
P(28≤ X≤94) = P(61≤ X≤94) = 99.7 %
This is so because the distribution is symmetric about the mean
