Question

Cherry trees in a certain orchard have heights that are normally distributed with mean = μ 106 inches and standard deviation = σ 17 .

(a) What proportion of trees are more than 109 inches tall?

(b) What proportion of trees are less than 92 inches tall?

(c) What is the probability that a randomly chosen tree is between 90 and 105 inches tall?

Round the answers to four decimal places.

Answer 1

Given : mean = 106 inches , standard deviation = 17 inches

a) P( x > 109) = 1 - P ( x < 109)

    = 1 - \phi [( x - mean)/standard deviation ]

= 1 - \phi [(109 - 106)/17]

= 1 - \phi (3/17)

= 1 - \phi (0.18)

= 1 - 0.5714

P ( x > 109) = 0.4286

Thus , 42.86 proportion of trees are more than 109 inches tall.

b) P( x <92) = \phi [(92 - mean)/standard deviation]

= \phi [(92 - 106)/17]

= \phi (-14/17)

= \phi ( - 0.82)

P( x< 92) = 0.2061

Thus, 20.61 proportion of trees are less than 92 inches tall.

c) P( 90 < x < 105) = P( x =105) - P( x = 90)

= \phi [(105 - mean)/standard deviation] - \phi [(90 - mean)/standard deviation]

= \phi [(105 - 106)/17] - \phi [(90 - 106)/17]

= \phi [(-1)/17] - \phi [(-16)/17]

= \phi (- 0.06) - \phi (- 0.94)

= 0.4761 - 0.1736

P(90< x < 105) = 0.3025

Thus , 30.25 proportion of trees are between 90 and 105 inches tall.

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