Question

a tree farm owner measure 27 trees in his garden centre. mean diamter of 10.4 inches and standard dev of 4.7 inches.
Draw normal model for tree farm
What size would you expect the central 95% of trees to be?(Diagram required)
What percent of trees should be less than an inch in diameter?(diagram required)
What percent of trees should be between 4.9 and 10.4 inches.(diagram required)

Answer 1

Question #71545, Math / Statistics and Probability

a tree farm owner measure 27 trees in his garden centre. mean diameter of 10.4 inches and standard dev of 4.7 inches.

Draw normal model for tree farm

Solution

Normal model:

What size would you expect the central 95% of trees to be? (Diagram required)

Solution

The z-scores associated with upper and lower 2.5% of data can be obtained from standard normal table or calculated using the technology (Excel function NORM.S.INV()). $z = \pm 1.96$

Lower endpoint = $\mu - z\sigma = 1.19$

Upper endpoint = $\mu + z\sigma = 19.61$

Central 95% of trees are expected to be between 1.19" and 19.61" diameter.

What percent of trees should be less than an inch in diameter?(diagram required)

Solution

The cumulative $p$ -value associated with the given data score can be calculated using $z$ -score and standard normal table, or using the technology (Excel function NORM.DIST()).

$p(x < 1) = 0.0228$

2.28% of trees are less than an inch in diameter.

What percent of trees should be between 4.9 and 10.4 inches.(diagram required)

Solution

$$p \left(x _ {1} < x < x _ {2}\right) = p \left(x < x _ {2}\right) - p \left(x < x _ {1}\right);$$ $$p (x < 1 0. 4) = 0. 5;$$ $$p (x < 4. 9 0) = 0. 1 2 1 0;$$ $$p (4. 9 0 < x < 1 0. 4) = 0. 5 - 0. 1 2 1 0 = 0. 3 7 9 0$$

37.90% of trees are between 4.9 and 10.4 inches.

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