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.

answer:   35 % is the probability of a late departure or an early arrival.

step-by-step explanation:

let l represents the late departure and e represents the early departure.

then, according to the question,

p(l) = 12 %,   p(e) = 27 % and = 4 %

and, we have to find out,

but, we know that,

which is the required equation that will use to find the probability of a late departure or an early departure.

therefore,   the probability of a late departure or an early arrival =   35 %

step-by-step explanation: hi!