Women’s heights are normally d
Women’s heights are normally distributed with mean 64 inches anda standard deviation of 2.5 inches.
(a) What percent of women are less than 60 inches tall?
(b) The requirement of all Navy pilots is between 62 inches and78 inches. What percent of women meet the Navy requirement?
(c) If the Navy changed its requirement to eliminate the”shortest” 5% of all women, what would the requirement be on thelower end of the scale?
Answer:
Solution :
Given that,
mean = 
= 64
standard deviation = 
= 2.5
A ) P( x < 60 )
P ( x – 
/) < ( 60 – 64 / 2.5 )
P ( z < – 4 / 2.5 )
P ( z < – 1.6 )
Using z table
= 0.0548
Probability = 0.0548
B ) P ( 62 < x < 58 )
P ( 62 – 60 / 2.5 ) < ( x – /) < ( 78 – 60 / 2.5)
P ( 2 / 2.5 < z < 18 / 2.5 )
P (0.8 < z < 7.2 )
P ( z < 7.2 ) – P ( z < 0.8)
Using z table
= 1 – 0.7881
= 0.2119
Probability = 0.2119
C ) Using standard normal table.
P(Z < z) = 5%
P(Z < z) = 0.05
P(Z < -1.645) = 0.05
z = -1.64
Using z-score formula,
x = z * 
+
x = -1.64 * 2.5 + 60
= 55.9
x = 55.9
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