Answer:
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
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-.52*86 + 1154 < x < 0.52*86 + 1154
1109.28 < x < 1198.72
Answer:
2 8/11 = b
Step-by-step explanation:
2b + 8 - 5b + 3 = -13 + 8b - 5
combine like terms
11 - 3b = -18 + 8b
add 3b
11 = -18 + 11b
add 18
29 = 11b
divide by 11
2 8/11 = b
