Based on the model N(11541154,8686) describing steer weights, what are the cutoff values for a) the highest 10% of the weights? a) the lowest 20% of the weights? b) the middle 40% of the weights? c) The cutoff weight for the highest 10% of the steer weights is nothing pounds.

a) x = 1225.68

b) x = 1081.76

c)  1109.28 < x < 1198.72

Step-by-step explanation:

Given:

- Th random variable X for steer weight follows a normal distribution:

                                    X~ N( 1154 , 86 )

Find:

a) the highest 10% of the weights?

b) the lowest 20% of the weights?

c) the middle 40% of the weights?

Solution:

a)

We will compute the corresponding Z-value for highest cut off 10%:

                                   Z @ 0.10 = 1.28

                                    Z = (x-u) / sd

Where,

u: Mean of the distribution.

s.d: Standard deviation of the distribution.

                                   1.28 = (x - 1154) / 86

                                      x = 1.28*86 + 1154

                                      x = 1225.68

b)

We will compute the corresponding Z-value for lowest cut off 20%:

                                   -Z @ 0.20 = -0.84

                                    Z = (x-u) / sd

                                   -0.84 = (x - 1154) / 86

                                      x = -0.84*86 + 1154

                                      x = 1081.76

c)

We will compute the corresponding Z-value for middle cut off 40%:

                                    Z @ 0.3 = -0.52

                                    Z @ 0.7 = 0.52

                                    [email protected] < x < [email protected]

                       -.52*86 + 1154 < x < 0.52*86 + 1154

                                  1109.28 < x < 1198.72