Respuesta :
Answer:
The middle 92% of all heights fall between 64.4 inches and 74.2 inches.
Step-by-step explanation:
Problems of normally distributed samples are 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.
In this problem, we have that:
[tex]\mu = 69.3, \sigma = 2.8[/tex]
Between what two values does that middle 92% of all heights fall?
The middle 92% falls from X when Z has a pvalue of 0.5 - 0.92/2 = 0.04 to X when Z has a pvalue of 0.5 + 0.92/2 = 0.96. So from the 4th percentile to the 96th percentile.
4th percentile
X when [tex]Z = -1.75[/tex]
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-1.75 = \frac{X - 69.3}{2.8}[/tex]
[tex]X - 69.3 = -1.75*2.8[/tex]
[tex]X = 64.4[/tex]
96th percentile
X when [tex]Z = 1.75[/tex]
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.75 = \frac{X - 69.3}{2.8}[/tex]
[tex]X - 69.3 = 1.75*2.8[/tex]
[tex]X = 74.2[/tex]
The middle 92% of all heights fall between 64.4 inches and 74.2 inches.
