10. [16 points] Suppose that the diameter at breast height (in inches) of trees

Question

10. [16 points] Suppose that the diameter at breast height (in inches) of trees of a certain type is normally distributed with f random. 8.5 and -25. Suppose that one tree of this type is selected at (a) Find the probability that the diameter of the tree is greater than 10 inches. 13 points] (b) Find the probability that the diameter of the tree is between 5 and 15 inches. [4 points] (c) Find the 25th percentile of the tree diameters. [3 points] (d) Between what two values are the middle 90% of tree diameters? [3 points] (e) If three trees are selected independently of each other, what is the probability that at least one of them has a diameter less than 10 inches? [3 points]

Explanation / Answer

mean = 8.5

sd = 2.5

a)

P(X > 10)

= P(z > (10 - 8.5)/2.5)

= P(z > 0.6)

= 0.2743

b)

P(5 < X < 15)

= P(X < 15) - P(X < 5)

= P(z < (15 - 8.5)/2.5) - P(z < (5 - 8.5)/2.5)

= P(z < 2.6) - P(z < -1.4)

= 0.9953 - 0.0808 .. (using standard table)

= 0.9146

c)

z-value for 25th percentile = -0.6745

x = mean + z*sigma

x = 8.5 + -0.6745*2.5

x = 6.8138 (required 25th percentile diameter)

d)

z-value of 90% = -/+ 1.6449

std. dev. = 2.5

SE = std.dev./sqrt(n) = 2.50000

ME = z*SE = 4.11213

Lower Limit = Mean - ME = 4.38787

Upper Limit = Mean + ME = 12.61213

Two values are (4.3879 , 12.6121 )

e)

P(X < 10) = 1- P(X > 10) = 1 - 0.2743 = 0.7257

here p = 0.7257

P(At least one) = 1 - (0.2743)^3 = 0.9794

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.