Question

3. Assume that oak trees have an average height of 90 feet with a standard

deviation of 14 feet. Their heights are normally distributed (i.e., μ = 90 and σ = 14).

A. Using a z table or online calculator, determine the percent of oak trees that are at least 106.50 feet tall. (Hint: You will need to start by converting 106.50 to a z score.)

B. Using a z table or online calculator, determine the percent of oak trees that are 83.95 feet or less.

C. Using your answers to A and B, what percent of oak trees’ heights are between 83.95 feet and 106.50 feet?

Answer 1

Solution:

Given: The heights oak trees are normally distributed with mean =  μ = 90 and standard deviation = σ = 14 .

Part A) Using a z table or online calculator, determine the percent of oak trees that are at least 106.50 feet tall.  

That is find:

P( X  ≥ 106.50) =..............?

Find z score for x = 106.50

Thus we get:

P( X  ≥ 106.50) = P( Z  ≥ 1.18 )

P( X  ≥ 106.50) = 1 - P( Z  < 1.18 )

Look in z table for z = 1.1 and 0.08 and find corresponding area.

P( Z< 1.18 ) = 0.8810

thus

P( X  ≥ 106.50) = 1 - P( Z  < 1.18 )

P( X  ≥ 106.50) = 1 - 0.8810

P( X  ≥ 106.50) = 0.1190

P( X  ≥ 106.50) = 11.90%

Part B)  Using a z table or online calculator, determine the percent of oak trees that are 83.95 feet or less.

That is find:

P( X ≤ 83.95 ) = ................?

thus

P( X ≤ 83.95 ) =P( Z ≤ -0.43)

Look in z table for z = -0.4 and 0.03 and find corresponding area.

P( Z< -0.43 ) =0.3336

thus

P( X ≤ 83.95 ) =P( Z ≤ -0.43)

P( X ≤ 83.95 ) = 0.3336

P( X ≤ 83.95 ) = 33.36%

Part C)  Using your answers to A and B, what percent of oak trees’ heights are between 83.95 feet and 106.50 feet?

That is find:

P( 83.95 < X < 106.50 ) = .............?

P( 83.95 < X < 106.50 ) = P( X < 106.50 ) - P( X < 83.95 )

P( 83.95 < X < 106.50 ) = P( Z < 1.18 ) - P( Z < -0.43)

From part A) we have : P( X < 106.50) = P( Z < 1.18) =0.8810 and from Part B) P( Z < -0.43) = 0.3336

P( 83.95 < X < 106.50 ) = 0.8810 - 0.3336

P( 83.95 < X < 106.50 ) = 0.5474

P( 83.95 < X < 106.50 ) = 54.74%