Respuesta :
Answer:
1. The minimum head breadth that will fit the clientele is of 4.41-in.
2. The maximum head breadth that will fit the clientele is of 7.19-in.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using 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.
Normally distributed with a mean of 5.8-in and a standard deviation of 0.8-in.
This means that [tex]\mu = 5.8, \sigma = 0.8[/tex]
1. What is the minimum head breadth that will fit the clientele?
The 4.1st percentile, that is, X when Z has a pvalue of 0.041, so X when Z = -1.74.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-1.74 = \frac{X - 5.8}{0.8}[/tex]
[tex]X - 5.8 = -1.74*0.8[/tex]
[tex]X = 4.41[/tex]
The minimum head breadth that will fit the clientele is of 4.41-in.
2. What is the maximum head breadth that will fit the clientele?
100 - 4.1 = 95.9th percentile, that is, X when Z has a pvalue of 0.959, so X when Z = 1.74.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.74 = \frac{X - 5.8}{0.8}[/tex]
[tex]X - 5.8 = 1.74*0.8[/tex]
[tex]X = 7.19[/tex]
The maximum head breadth that will fit the clientele is of 7.19-in.
