Question

Horses in a stable have a mean weight of 950 pounds with a standard deviation of 77 pounds. Weights of horses follow the normal distribution. One horse is selected at random.

a. What is the probability that the horse weighs less than 900 pounds?

b. What is the probability that the horse weighs more than 1,100 pounds?

c. What is the probability that the horse weighs between 900 and 1,100 pounds?

d. What weight is the 90th percentile? (round to the nearest pound)

Answer 1

a)

Here, μ = 950, σ = 77 and x = 900. We need to compute P(X <= 900). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (900 - 950)/77 = -0.65

Therefore,
P(X <= 900) = P(z <= (900 - 950)/77)
= P(z <= -0.65)
= 0.2578

b)

Here, μ = 950, σ = 77 and x = 1100. We need to compute P(X >= 1100). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (1100 - 950)/77 = 1.95

Therefore,
P(X >= 1100) = P(z <= (1100 - 950)/77)
= P(z >= 1.95)
= 1 - 0.9744 = 0.0256

c)

Here, μ = 950, σ = 77, x1 = 900 and x2 = 1100. We need to compute P(900<= X <= 1100). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (900 - 950)/77 = -0.65
z2 = (1100 - 950)/77 = 1.95

Therefore, we get
P(900 <= X <= 1100) = P((1100 - 950)/77) <= z <= (1100 - 950)/77)
= P(-0.65 <= z <= 1.95) = P(z <= 1.95) - P(z <= -0.65)
= 0.9744 - 0.2578
= 0.7166

d)

z value at 90 = 1.28

z = (x - mena)/s

1.28 = (x - 950)/77
x = 77 * 1.28+ 950
x = 1049