Respuesta :
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([tex]\mu=0,\sigma^{2} = 5.9^{2}[/tex])
The z score probability distribution is given by ;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = mean residual = 0 cm
[tex]\sigma[/tex] = 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( [tex]\frac{X-\mu}{\sigma}[/tex] > [tex]\frac{8-0}{5.9}[/tex] ) = P(Z > 1.36) = 1 - P(Z [tex]\leq[/tex] 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.
Using the normal distribution, it is found that 8.69% of the residuals are greater than 8cm.
-------------------------
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- The Z-score measures how many standard deviations the measure is from the mean, and has a p-value associated.
- The p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
-------------------------
- Mean of 0cm means that [tex]\mu = 0[/tex]
- Standard deviation of 5.9cm means that [tex]\sigma = 5.9[/tex]
The proportion above 8 cm is 1 subtracted by the p-value of Z when X = 8, thus:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{8 - 0}{5.9}[/tex]
[tex]Z = 1.36[/tex]
[tex]Z = 1.36[/tex] has a p-value of 0.9131.
1 - 0.9131 = 0.0869
As a percentage, 0.0869 x 100% = 8.69%.
8.69% of the residuals are greater than 8cm.
A similar problem is given at https://brainly.com/question/16080952
