Question

0. Foot length (in inches) of a randomly chosen adult male is a normal random variable with a mean of 11 and standard deviation of 1.5. Use this information and the z-table to answer the following questions a-j.

   1. On average, what is a randomly chosen adult male’s foot length?

   2. Let X denote the foot length of a randomly chosen adult male. What is the distribution of X?

   3. What is the probability that a randomly chosen adult male’s foot length will be less than 10?

   4. What is the probability that a randomly chosen adult male’s foot length will be less than 14?

   5. What is the probability that a randomly chosen adult male’s foot length will be between 10 and 13?

   6. What is the probability that a randomly chosen adult male’s foot length will be between 11 and 14?

   7. What is the probability that a randomly chosen adult male’s foot length is greater than 12?

   8. What is the probability that a randomly chosen adult male’s foot length is greater than 13?

   9. Find the foot length “c” such that probability of a randomly chosen adult male’s foot length greater than c is 0.5.

   10. Find the foot length “r” such that probability of a randomly chosen adult male’s foot length less than r is 0.2.

Answer 1

Part a)

X ~ N ( µ = 11 , σ = 1.5 )

Part b)

P ( X < 10 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 10 - 11 ) / 1.5
Z = -0.6667
P ( ( X - µ ) / σ ) < ( 10 - 11 ) / 1.5 )
P ( X < 10 ) = P ( Z < -0.6667 )
P ( X < 10 ) = 0.2525

Part c)

X ~ N ( µ = 11 , σ = 1.5 )
P ( X < 14 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 14 - 11 ) / 1.5
Z = 2
P ( ( X - µ ) / σ ) < ( 14 - 11 ) / 1.5 )
P ( X < 14 ) = P ( Z < 2 )
P ( X < 14 ) = 0.9772

Part d)

X ~ N ( µ = 11 , σ = 1.5 )
P ( 10 < X < 13 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 10 - 11 ) / 1.5
Z = -0.6667
Z = ( 13 - 11 ) / 1.5
Z = 1.3333
P ( -0.67 < Z < 1.33 )
P ( 10 < X < 13 ) = P ( Z < 1.33 ) - P ( Z < -0.67 )
P ( 10 < X < 13 ) = 0.9088 - 0.2525
P ( 10 < X < 13 ) = 0.6563

Part e)

X ~ N ( µ = 11 , σ = 1.5 )
P ( 11 < X < 14 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 11 - 11 ) / 1.5
Z = 0
Z = ( 14 - 11 ) / 1.5
Z = 2
P ( 0 < Z < 2 )
P ( 11 < X < 14 ) = P ( Z < 2 ) - P ( Z < 0 )
P ( 11 < X < 14 ) = 0.9772 - 0.5
P ( 11 < X < 14 ) = 0.4772

Part f)

X ~ N ( µ = 11 , σ = 1.5 )
P ( X > 12 ) = 1 - P ( X < 12 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 12 - 11 ) / 1.5
Z = 0.6667
P ( ( X - µ ) / σ ) > ( 12 - 11 ) / 1.5 )
P ( Z > 0.6667 )
P ( X > 12 ) = 1 - P ( Z < 0.6667 )
P ( X > 12 ) = 1 - 0.7475
P ( X > 12 ) = 0.2525

Part g)

X ~ N ( µ = 11 , σ = 1.5 )
P ( X > 13 ) = 1 - P ( X < 13 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 13 - 11 ) / 1.5
Z = 1.3333
P ( ( X - µ ) / σ ) > ( 13 - 11 ) / 1.5 )
P ( Z > 1.3333 )
P ( X > 13 ) = 1 - P ( Z < 1.3333 )
P ( X > 13 ) = 1 - 0.9088
P ( X > 13 ) = 0.0912

Part h)

X ~ N ( µ = 11 , σ = 1.5 )
P ( X > x ) = 1 - P ( X < x ) = 1 - 0.5 = 0.5
to find the value of x
Looking for the probability 0.5 in standard normal table to calculate critical value Z = 0
Z = ( X - µ ) / σ
0 = ( X - 11 ) / 1.5
X = 11
P ( X > 11 ) = 0.5

Part i)

X ~ N ( µ = 11 , σ = 1.5 )
P ( X < x ) = 20% = 0.2
to find the value of x
Looking for the probability 0.2 in standard normal table to calculate critical value Z = -0.8416
Z = ( X - µ ) / σ
-0.8416 = ( X - 11 ) / 1.5
X = 9.7376
P ( X < 9.7376 ) = 0.2