Quadrilateral ABCD in the figure below represents a scaled-down model of a walkway around a historic site. Quadrilateral EFGH re
presents the actual walkway. http://assets.openstudy.com/updates/attachments/4fb5807de4b05565342b5f98-bbabycakes6921-1337295128659-image0024e9d8a3f.jpg ABCD is similar to EFGH.. . What is the total length, in feet, of the actual walkway?
2 answers:
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Answer:
μ=23.710740....
σ=3.592592....
Step-by-step explanation:
At first, we are told that it´s a normal distribution and the following:
P(x≤15.34)=0.1
P(x≤16.31)=0.2
We start looking for those values (0.1 and 0.2) and the probabilities that gives us those values in a normal standard distribution. There´s some tools that´ll help us out but i´ll be using a chart with the values of some probabilities in a normal standard distribution (That it´s attached).
The probability of P(z≤ a.b) where "a" is the whole part of a number and "b" the decimal part is in the coordinates (a,b). For example for our problem:
P(z≤-2.33)=0.099≈0.1
P(z≤-2.06)=0.197≈0.2
To find a probability of a normal (not standard) distribution we use a method called Normalize that proceeds:
P(x≤a)=P( (x-μ)/σ ≤ (a-μ)/σ )
Where μ is the Mean of the data and σ is the Standard deviation. P( (x-μ)/σ ≤ (a-μ)/σ ) is now part of a normal standard distribution and we are able to look for it in the chart.
We will use it to find the Mean and the STD for our distribution.
P(x≤15.34)=P(z≤-2.33)≈0.1
P(x≤16.31)=P(z≤-2.06)≈0.2
so we have the equations:
(15.34 - μ)/σ= -2.33 and (16.31 - μ)/σ= -2.06
And we solve them for μ and σ:
(15.34 - μ)/-2.33= σ and (16.31 - μ)/-2.06= σ
(15.34 - μ)/-2.33 = (16.31 - μ)/-2.06
2.06*(15.34 - μ) = 2.33(16.31 - μ)
2.33μ - 2.06μ = 2.33(16.31) - 2.06(15.34)
μ= 6.4019 / 0.27
μ= 23.710740...
With this value we find σ in one equation:
(15.34 - (23.710740...)/σ = -2.33
(15.34 - (23.710740...)/-2.33 = σ
σ=3.592592...
We finally have our answers:
μ=23.710740....
σ=3.592592....
