Question

QUESTION 5

Use your TI83 (or Excel):

A normally distributed population has a mean of 72 and a standard deviation of 14. Determine the probability that a random sample of size 35 has an average of less than 73.

Round to four decimal places.

QUESTION 6

A normally distributed population has a mean of 71 and a standard deviation of 15. Determine the probability that a random sample of size 26 has an average greater than 74.

Round to four decimal places.

QUESTION 7

A normally distributed population has a mean of 76 and a standard deviation of 14. Determine the probability that a random sample of size 25 has an average of less than 74.

Round to four decimal places.

QUESTION 8

Use your TI83 (or Excel):

A normally distributed population has a mean of 76 and a standard deviation of 16. Determine the probability that a random sample of size 36 has an average greater than 74.

Round to four decimal places.

Answer 1

Question 5

X ~ N ( µ = 72 , σ = 14 )
P ( X < 73 )
Standardizing the value
Z = ( X - µ ) / (σ/√(n)
Z = ( 73 - 72 ) / ( 14 / √35 )
Z = 0.4226

Using excel  NORMSDIST(0.4226)

µX̅ = µ = 72
σX̅ = σ / √ (n) = 14/√35 = 2.3664

In TI 84

Question 6

X ~ N ( µ = 71 , σ = 15 )
P ( X > 74 ) = 1 - P ( X < 74 )
Standardizing the value
Z = ( X - µ ) / ( σ / √(n))
Z = ( 74 - 71 ) / ( 15 / √ ( 26 ) )
Z = 1.0198

Using Excel  1-NORMSDIST(1.0198)

In TI 84

µX̅ = µ = 71
σX̅ = σ / √ (n) = 15/√26 = 2.9417

Question 7

X ~ N ( µ = 76 , σ = 14 )
P ( X < 74 )
Standardizing the value
Z = ( X - µ ) / (σ/√(n)
Z = ( 74 - 76 ) / ( 14 / √25 )
Z = -0.7143

Excel formula NORMSDIST(-0.7143)

In TI 84

µX̅ = µ = 76
σX̅ = σ / √ (n) = 14/√25 = 2.8

Question 8

X ~ N ( µ = 76 , σ = 16 )
P ( X > 74 ) = 1 - P ( X < 74 )
Standardizing the value
Z = ( X - µ ) / ( σ / √(n))
Z = ( 74 - 76 ) / ( 16 / √ ( 36 ) )
Z = -0.75

Excel formula 1-NORMSDIST(-0.75)

In TI 84

µX̅ = µ = 76
σX̅ = σ / √ (n) = 16/√36 = 2.6667