6. Heights of Kookooberry trees are normally distributed with a mean of 7.5 feet

Question

6. Heights of Kookooberry trees are normally distributed with a mean of 7.5 feet and a standard deviation of 1.2 feet. a. If I randomly find a Kookooberry tree while taking a nature walk, what's the b. If I randomly find a Kookooberry tree while taking a nature walk, what's the c. Mrs. Wilson is landscaping her yard and wants to find several Kookooberry probability that it will be over 9 feet tall? probability that it will be between 6 and 7 feet tall? trees under 5 feet tall, what z-score represents the cut-off height for the trees she is seeking?

Explanation / Answer

Question 6

We are given heights are normally distributed with mean = 7.5 and SD = 1.2.

Part a

We have to find P(X>9)

P(X>9) = 1 – P(X<9)

Z = (X – mean) / SD

Z = (9 – 7.5) / 1.2

Z = 1.25

P(Z<1.25) = P(X<9) = 0.89435

(by using z-table)

P(X>9) = 1 – P(X<9)

P(X>9) = 1 – 0.89435

P(X>9) = 0.10565

Required probability = 0.10565

Part b

We have to find P(6<X<7)

P(6<X<7) = P(X<7) – P(X<6)

For X = 7, z-score is given as below:

Z = (X – mean) / SD

Z = (7 – 7.5)/1.2

Z = -0.41667

P(X<7) = P(Z< -0.41667) = 0.338461

Now, we have to find z-score for X = 6

Z = (6 – 7.5) / 1.2

Z = -1.25

P(X<6) = P(Z< -1.25) = 0.10565

P(6<X<7) = P(X<7) – P(X<6)

P(6<X<7) = 0.338461 - 0.10565

P(6<X<7) = 0.232811

Required probability = 0.232811

Part c

Here, we have to find z score for X = 5 as a cut-off height.

Z = (X – mean) / SD

Z = (5 – 7.5) / 1.2

Z = -2.08333

Required z-score = -2.08333

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