Assume that the distribution of residuals is approximately normal with mean 0cm and standard deviation 5.9cm . What percent of the residuals are greater than 8cm ? Justify your answer

8.7% of the residuals are greater than 8 cm.

Step-by-step explanation:

We are given that the distribution of residuals is approximately normal with mean 0 cm and standard deviation 5.9 cm.

Let X = distribution of residuals

So, X ~ N()

The z score probability distribution is given by ;

           Z =  ~ N(0,1)

where, = mean residual = 0 cm

            = standard deviation = 5.9 cm

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

So, percent of the residuals that are greater than 8 cm is given by = P(X > 8 cm)

     P(X > 8 cm) = P( > ) = P(Z > 1.36) = 1 - P(Z 1.36)

                                                   = 1 - 0.9131 = 0.0869  or 8.7%

The above probability is calculated using z table by looking at value of x = 1.36 in the z table which have an area of 0.9131.

Therefore, 8.7% of the residuals are greater than 8 cm.

geometry: triangles

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lessons

answers archive

call the larger triangle abc

then the smaller triangle cde is similar to abc (all angles are equal, aaa).  

ab is the side that is 36, and we wish to find de, the side corresponding to ab

area_cde = area_abc - area_cde --> area_abc / area_cde = 2

let x=length of de

ratio of areas = (ratio of sides)^2

+2+=+%2836%2fx%29%5e2+

+2+=+1296%2fx%5e2+

++x%5e2+=+1296%2f2+=+648+

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