Question

Based on the model N(1156,85​) describing steer​ weights, what are the cutoff values for

​a) the highest​ 10% of the​ weights?

​b) the lowest​ 20% of the​ weights?

​c) the middle​ 40% of the​ weights?

​a) The cutoff weight for the highest​ 10% of the steer weights is __lbs.

​(Round to one decimal place as​ needed.)

​b) The cutoff weight for the lowest​ 20% of the steer weights is __lbs

​(Round to one decimal place as​ needed.)

​c) The cutoff for the middle​ 40% of the steer weights are __ and __obs

Answer 1

Given that, mean (μ) = 1156 and

standard deviation = 85

X ~ N(1156, 85)

a) We want to find the value of x such that, P(X > x) = 0.10

Therefore, the cutoff weight for the highest​ 10% of the steer weights is 1264.8 lbs.

b) We want to find the value of x such that, P(X < x) = 0.20

Therefore, the cutoff weight for the lowest​ 20% of the steer weights is 1084.6 lbs.

c) We want to find, the value of x1 and x2 such that,

P(x1 < X < x2) = 0.40

First we find z-score such that,

P(-z < Z < z) = 0.40

=> 2 * P(Z < z) - 1 = 0.40

=> 2 * P(Z < z) = 1.40

=> P(Z < z) = 0.7

Using standard normal z-table we get z-score corresponding probability of 0.7 is, z = 0.52

For z = -0.52

x1 = (-0.52 * 85) + 1156 = -44.2 + 1156 = 1111.8

For z = 0.52

x2 = (0.52 * 85) + 1156 = 44.2 + 1156 = 1200.2

Therefore, the cutoff for the middle​ 40% of the steer weights are 1111.8 lbs and 1200.2 lbs