Respuesta :
Answer:
a) 3.3352 inches.
b) 8.2648 inches.
Step-by-step explanation:
When the distribution is normal, we use the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore 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. After finding the Z-score, we look at the z-score table and find the p-value 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. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this question:
[tex]\mu = 5.8, \sigma = 1.2[/tex]
A. What is the minimum head breadth that will fit the clientele?
This is the 2nd percentile, which is X when Z has a pvalue of 0.02. So X when Z = -2.054.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-2.054 = \frac{X - 5.8}{1.2}[/tex]
[tex]X - 5.8 = -2.054*1.2[/tex]
[tex]X = 3.3352[/tex]
So the minimum head breadth that will fit the clientele is 3.3352 inches.
B. What is the maximum head breadth that will fit the clientele?
The 100-2 = 98th percentile, which is X when Z has a pvalue of 0.98. So X when Z = 2.054.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]2.054 = \frac{X - 5.8}{1.2}[/tex]
[tex]X - 5.8 = 2.054*1.2[/tex]
[tex]X = 8.2648[/tex]
So the maximum head breadth that will fit the clientele is 8.2648 inches.
