Question

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

Answer 1

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($\mu=0,\sigma^{2} = 5.9^{2}$)

The z score probability distribution is given by ;

           Z = $\frac{X-\mu}{\sigma}$  ~ N(0,1)

where, $\mu$ = mean residual = 0 cm

            $\sigma$ = 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( $\frac{X-\mu}{\sigma}$ > $\frac{8-0}{5.9}$ ) = P(Z > 1.36) = 1 - P(Z $\leq$ 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.