A citrus farmer who grows mandarin oranges finds that the diameters of mandarin

Question

A citrus farmer who grows mandarin oranges finds that the diameters of mandarin oranges harvested on his farm follow a normal distribution with a mean diameter of 5.85 cm and a standard deviation of 0.24 cm.

a. Find the probability that a randomly selected mandarin orange from this farm has a diameter larger than 6.0 cm. (Answer to 2 decimal places.)

b. The middle 20% of mandarin oranges from this farm have diameters between and . (Answer to 2 decimal places.)

c. Find the 90th percentile for the diameters of mandarin oranges, and interpret it in a complete sentence. (to 2 decimal places).

d. The middle 40% of mandarin oranges from this farm are between and . (Answer to 2 decimal places.)

e. Find the 16th percentile. To 2 decimal places

Explanation / Answer

Mean = 5.85 cm

Standard deviation = 0.24 cm

P(X < A) = P(Z < (A - mean)/standard deviation

a) P(X > 6) = 1 - P(X < 6)

= 1 - P(Z < (6 - 5.85)/0.24)

= 1 - P(Z < 0.625)

= 1 - 0.7340

= 0.2660

b) Let A nad B indicate the intervals.

P(X < A) = 0.4

P(Z < (A - 5.85)/0.24) = 0.4

(A - 5.85)/0.24 = -0.25

A = 5.85 - 0.06 = 5.79

B = 5.85 + 0.06 = 5.91

Middle 20% will have diameter between 5.79 cm and 5.91 cm

c) Let M indicate the 90th percentile

P(X < M ) = 0.9

P(Z < (M - 5.85)/0.24) = 0.9

(M - 5.85)/0.24 = 1.28

M = 6.16 cm

90% of the mandarin oranges will have a mean diameter less than 6.16 cm

d) Let P nad Q indicate the intervals.

P(X < A) = 0.3

P(Z < (A - 5.85)/0.24) = 0.3

(A - 5.85)/0.24 = -0.52

A = 5.85 - 0.12 = 5.73

B = 5.85 + 0.12 = 5.97

Middle 40% will have diameter between 5.73 cm and 5.97 cm

e)

Let N indicate the 16th percentile

P(X < N ) = 0.16

P(Z < (N - 5.85)/0.24) = 0.16

(N - 5.85)/0.24 = -0.99

N = 5.61 cm

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