Question

The heights of pecan trees are normally distributed with a mean of 10 feet and a standard deviation of 2 feet. Show all work. Just the answer, without supporting work, will receive no credit. (a) What is the probability that a randomly selected pecan tree is between 9 and 12 feet tall? (round the answer to 4 decimal places) (b) Find the 75th percentile of the pecan tree height distribution. (round the answer to 2 decimal places) (c) To get the answers for part (a) and part (b), what technology did you use? If an online applet was used, list the URL and describe the steps. If a calculator or Excel was used, write out the function

Answer 1

Solution:

   We are given that: The heights of pecan trees are normally distributed with a mean of 10 feet and a standard deviation of 2 feet.

That is: Mean = and Standard Deviation =

Part a) What is the probability that a randomly selected pecan tree is between 9 and 12 feet tall?

That is:

P( 9 < X < 12 ) =..........?

find z scores for x = 12 and for x = 9:

Thus we get:

P( 9 < X < 12 ) =P( -0.50 < Z < 1.00)

P( 9 < X < 12 ) =P( Z < 1.00) - P( Z < -0.50)

Look in z table for z = 1.0 and 0.00 and find area as well as look for z = -0.5 and 0.00 and find area.

P( Z < -0.50) = 0.3085

P( Z < 1.00) = 0.8413

Thus

P( 9 < X < 12 ) =P( Z < 1.00) - P( Z < -0.50)

P( 9 < X < 12 ) = 0.8413 - 0.3085

P( 9 < X < 12 ) = 0.5328

We also can use excel:

=NORM.DIST( x , mean , SD , cumulative )

Thus to find P( 9 < X < 12) , we use following step:

=NORM.DIST( 12 , 10 , 2 , TRUE) - NORM.DIST( 9 , 10 , 2 , TRUE)

then press Enter

which gives answer

=0.532807

=0.5328

Part b) Find the 75th percentile of the pecan tree height distribution.

We have to find x value such that:

P( X < x ) = 75%

P( X < x ) = 0.75

Thus find z value such that:
P( Z < z ) = 0.75

Look in z table for area = 0.7500 or its closest area and find its corresponding z value:

Area = 0.7486 is closest to 0.7500

which corresponds to 0.6 and 0.07

thus z = 0.67

Thus we use following formula:

We use following Excel command:

=NORMINV( Probability , Mean , SD )

=NORMINV( 0.75 , 10 , 2 )

=11.349

=11.35

In excel we get exact answer for given area.

Part c) To get the answers for part (a) and part (b), what technology did you use?

We used Excel:

a) Excel command for probability:

=NORM.DIST( x , mean , SD , cumulative )

=NORM.DIST( 12 , 10 , 2 , TRUE) - NORM.DIST( 9 , 10 , 2 , TRUE)

=0.5328

b) Excel command for x value:

=NORMINV( Probability , Mean , SD )

=NORMINV( 0.75 , 10 , 2 )

=11.349

=11.35