Answer:
a) 6.68% of heights less than 150 centimeters
b) 58.65% of heights between 160 centimeters and 180 centimeters
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, we have that:
[tex]\mu = 162, \sigma = 8[/tex]
a) The percentage of heights less than 150 centimeters
We have to find the pvalue of Z when X = 150. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{150 - 162}{8}[/tex]
[tex]Z = -1.5[/tex]
[tex]Z = -1.5[/tex] has a pvalue of 0.0668
6.68% of heights less than 150 centimeters
b) The percentage of heights between 160 centimeters and 180 centimeters
We have to find the pvalue of Z when X = 180 subtracted by the pvalue of Z when X = 160.
X = 180
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{180 - 162}{8}[/tex]
[tex]Z = 2.25[/tex]
[tex]Z = 2.25[/tex] has a pvalue of 0.9878
X = 160
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{160 - 162}{8}[/tex]
[tex]Z = -0.25[/tex]
[tex]Z = -0.25[/tex] has a pvalue of 0.4013
0.9878 - 0.4013 = 0.5865
58.65% of heights between 160 centimeters and 180 centimeters
