Question

Student scores on Professor Combs' Stats final exam are normally distributed with a mean of 77 and a standard deviation of 7.5 Find the probability of the following: **(use 4 decimal places)** a.) The probability that one student chosen at random scores above an 82. b.) The probability that 20 students chosen at random have a mean score above an 82. c.) The probability that one student chosen at random scores between a 72 and an 82. d.) The probability that 20 students chosen at random have a mean score between a 72 and an 82.

q2. World class marathon runners are known to run that distance (26.2 miles) in an average of 143 minutes with a standard deviation of 13 minutes.

If we sampled a group of world class runners from a particular race, find the probability of the following:

**(use 4 decimal places)**

a.) The probability that one runner chosen at random finishes the race in less than 137 minutes.

b.) The probability that 10 runners chosen at random have an average finish time of less than 137 minutes.

c.) The probability that 50 runners chosen at random have an average finish time of less than 137 minutes.

Answer 1

Question 1

Part a)
X ~ N ( µ = 77 , σ = 7.5 )
P ( X > 82 ) = 1 - P ( X < 82 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 82 - 77 ) / 7.5
Z = 0.6667
P ( ( X - µ ) / σ ) > ( 82 - 77 ) / 7.5 )
P ( Z > 0.6667 )
P ( X > 82 ) = 1 - P ( Z < 0.6667 )
P ( X > 82 ) = 1 - 0.7475
P ( X > 82 ) = 0.2525

Part b)
X ~ N ( µ = 77 , σ = 7.5 )
P ( X > 82 ) = 1 - P ( X < 82 )
Standardizing the value
Z = ( X - µ ) / ( σ / √(n))
Z = ( 82 - 77 ) / ( 7.5 / √ ( 20 ) )
Z = 2.9814
P ( ( X - µ ) / ( σ / √ (n)) > ( 82 - 77 ) / ( 7.5 / √(20) )
P ( Z > 2.98 )
P ( X̅ > 82 ) = 1 - P ( Z < 2.98 )
P ( X̅ > 82 ) = 1 - 0.9986
P ( X̅ > 82 ) = 0.0014

Part c)
X ~ N ( µ = 77 , σ = 7.5 )
P ( 72 < X < 82 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 72 - 77 ) / 7.5
Z = -0.6667
Z = ( 82 - 77 ) / 7.5
Z = 0.6667
P ( -0.67 < Z < 0.67 )
P ( 72 < X < 82 ) = P ( Z < 0.67 ) - P ( Z < -0.67 )
P ( 72 < X < 82 ) = 0.7475 - 0.2525
P ( 72 < X < 82 ) = 0.495

Part d)
X ~ N ( µ = 77 , σ = 7.5 )
P ( 72 < X < 82 )
Standardizing the value
Z = ( X - µ ) / ( σ / √(n))
Z = ( 72 - 77 ) / ( 7.5 / √(20))
Z = -2.9814
Z = ( 82 - 77 ) / ( 7.5 / √(20))
Z = 2.9814
P ( -2.98 < Z < 2.98 )
P ( 72 < X̅ < 82 ) = P ( Z < 2.98 ) - P ( Z < -2.98 )
P ( 72 < X̅ < 82 ) = 0.9986 - 0.0014
P ( 72 < X̅ < 82 ) = 0.9971

Question 2

Part a)
X ~ N ( µ = 143 , σ = 13 )
P ( X < 137 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 137 - 143 ) / 13
Z = -0.4615
P ( ( X - µ ) / σ ) < ( 137 - 143 ) / 13 )
P ( X < 137 ) = P ( Z < -0.4615 )
P ( X < 137 ) = 0.3222

Part b)
X ~ N ( µ = 143 , σ = 13 )
P ( X < 137 )
Standardizing the value
Z = ( X - µ ) / (σ/√(n)
Z = ( 137 - 143 ) / ( 13 / √10 )
Z = -1.4595
P ( ( X - µ ) / ( σ/√(n)) < ( 137 - 143 ) / ( 13 / √(10) )
P ( X < 137 ) = P ( Z < -1.46 )
P ( X̅ < 137 ) = 0.0722

Part c)
X ~ N ( µ = 143 , σ = 13 )
P ( X < 137 )
Standardizing the value
Z = ( X - µ ) / (σ/√(n)
Z = ( 137 - 143 ) / ( 13 / √50 )
Z = -3.2636
P ( ( X - µ ) / ( σ/√(n)) < ( 137 - 143 ) / ( 13 / √(50) )
P ( X < 137 ) = P ( Z < -3.26 )
P ( X̅ < 137 ) = 0.0006