Respuesta :
The heights of women aged 20 to 29 are approximately Normal with mean 64 inches and standard deviation 2.7 inches. Men the same age have mean height 69.3 inches with standard deviation 2.8 inches.
Answer:
a) [tex]Z = 4.4[/tex]
b) [tex]Z = 0.38[/tex]
c) The woman is relatively taller.
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.
a) What is the z-score for a woman who is 6 feet tall? Please use 2 decimal places.
The heights of women aged 20 to 29 are approximately Normal with mean 61 inches and standard deviation 2.5 inches, which means that [tex]\mu = 61, \sigma = 2.5[/tex].
The mean and the standard deviation are in inches, so X is also must be in inches. Each feet has 12 inches. So X = 6*12 = 72 inches.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{72 - 61}{2.5}[/tex]
[tex]Z = 4.4[/tex]
(b) What is the z-score for a man who is 6 feet tall? Please use 2 decimal places.
Men the same age have mean height 71 inches with standard deviation 2.6 inches, which means that [tex]\mu = 71, \sigma = 2.6[/tex]
The mean and the standard deviation are in inches, so X is also must be in inches. Each feet has 12 inches. So X = 6*12 = 72 inches.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{72 - 71}{2.6}[/tex]
[tex]Z = 0.38[/tex]
(c) Who is relatively taller?
The woman has the higher z-score, so she is relatively taller.
