The heights of women aged 20 to 29 are approximately normal with mean 64 inches

Question

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. a) What is the z-scores for a woman 6 feet tall? b) What is the z-scores for a man 6 feet tall? c) What information do the z-scores give that the actual heights do not? d) What percent of men are shorter than 5 feet, 5 inches tall? e) What percent of women are over 6 feet, 1 inch tall? f) Find the interquartile range for the height of women aged 20-29.

Explanation / Answer

a)

x = 6 ft = 72 in

Hence,

z = (x-u)/sigma = (72-64)/2.7 = 2.96296 [ANSWER]

*****************

b)

z = (x-u)/sigma = (72-69.3)/2.8 = 0.964285 [ANSWER]

*******************

c)

It lets you compare the raw score to the other scores in the population. The actual high does not do this.

*******************

d)

5'5" = 65 in.

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    65      
u = mean =    69.3      
          
s = standard deviation =    2.8      
          
Thus,          
          
z = (x - u) / s =    -1.535714286      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -1.535714286   ) =    0.062304235 = 6.23% [ANSWER]

*******************

e)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    73      
u = mean =    64      
          
s = standard deviation =    2.7      
          
Thus,          
          
z = (x - u) / s =    3.333333333      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   3.333333333   ) =    0.00042906 = 0.0429% [ANSWER]

*********************

f)

We are talking about the middle 50%.

As the middle area is          
          
Middle Area = P(x1<x<x2) =    0.5      
          
Then the left tailed area of the left endpoint is          
          
P(x<x1) = (1-P(x1<x<x2))/2 =    0.25      
          
Thus, the z score corresponding to the left endpoint, by table/technology, is          
          
z1 =    -0.67448975      
By symmetry,          
z2 =    0.67448975      
          
As          
          
u = mean =    64      
s = standard deviation =    2.7      
          
Then          
          
x1 = u + z1*s =    62.17887767   [Q1]  
x2 = u + z2*s =    65.82112233   [Q3]  

Hence,

IQR = Q3 - Q1 = 3.642244651 [ANSWER]

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.