Question

Getting Startedaleksyunisex/ 10 U-INSIRFJOP3JH-likypekuw5CXnZGOLVS 9076Activities - IHX Nation,Tree heights: Cherry trees in a certain orchard have heights that are normally distributed with mean = 120 inches andstandard deviation o=15.(a) What proportion of trees are more than 127 Inches tall?E(b) What proportion of trees are less than 104 inches tall?(c) What is the probability that a randomly chosen tree is between 91 and 112 inches tall?Round the answers to four decimal places.Part 1 of 3alThe proportion of trees that are more than 127 inches tall isPart 2 of 3The proportion of trees that are less than 104 inches tall isPart 3 of 3Save For LaterSubmit AssignmentCheck Answer

Answer 1

Step 1 : Let's review the information provided to us to answer the questions correctly:

• Mean of height of Cherry trees in a certain orchard = 120 inches

,

• Standard deviation = 15 inches

Step 2: Let's calculate the proportion of trees that are more than 127 Inches tall

We need to find the z-score for 127, this way:

• z-score = (127 - 120)/15

,

• z-score = 7/15

,

• z-score = 0.467

Using the z-table for calculating the proportion, we have:

P-value from Z-Table:

P( x < 127) = 0.67963

But we need to know the proportion of trees higher than 127 inches, therefore:

P( x > 127) = 1 - P( x < 127) = 1 - 0.67963 = 0.32037

The answer is 0.3204 (rounding to four decimal places) or 32.04%

Step 3: Let's calculate the proportion of trees that are less than 104 Inches tall

We need to find the z-score for 104, this way:

• z-score = (104 - 120)/15

,

• z-score = -16/15

,

• z-score = - 1.06667

Using the z-table for calculating the proportion, we have:

P-value from Z-Table:

P( x < 104) = 0.14306

The answer is 0.1431 (rounding to four decimal places) or 14.31%

Step 4: Let's calculate the proportion that a randomly chosen tree is between 91 and 112 inches tall.

We need to find the z-score for 91, this way:

• z-score = (91 - 120)/15

,

• z-score = -29/15

,

• z-score = -1.93333

We need to find the z-score for 112, this way:

z-score = (112 - 120)/15

z-score = -8/15

z-score = - 0.53333

Using the z-table for calculating the proportion, we have:

P-value from Z-Table:

P(x>-0.533330) = 0.7031

P(x<-1.93333) = 0.026598

Therefore:

P(x<-1.93333 or x>-0.533330) = 0.7297

But we need to calculate the proportion of trees that are between 91 and 112

P(-1.93333-0.533330) = 1 - 0.7297 = 0.2703

The answer is 0.2703 (rounding to four decimal places) or 27.03%

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