Respuesta :
Answer:
You should expect to find the middle 98% of most head breadths between 3.34 in and 8.46 in.
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 = 5.9, \sigma = 1.1[/tex]
In what range would you expect to find the middle 98% of most head breadths?
From the: 50 - (98/2) = 1st percentile.
To the: 50 + (98/2) = 99th percentile.
1st percentile:
X when Z has a pvalue of 0.01. So X when Z = -2.327.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-2.327 = \frac{X - 5.9}{1.1}[/tex]
[tex]X - 5.9 = -2.327*1.1[/tex]
[tex]X = 3.34[/tex]
99th percentile:
X when Z has a pvalue of 0.99. So X when Z = 2.327.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]2.327 = \frac{X - 5.9}{1.1}[/tex]
[tex]X - 5.9 = 2.327*1.1[/tex]
[tex]X = 8.46[/tex]
You should expect to find the middle 98% of most head breadths between 3.34 in and 8.46 in.
