Respuesta :
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
x@0.3 < x < x@0.7
-.52*86 + 1154 < x < 0.52*86 + 1154
1109.28 < x < 1198.72
Using the normal distribution, it is found that:
a) The cutoff for the highest 10% of pounds is of 1264.
b) The cutoff for the lowest 20% of weights is of 1082.
c) The cutoff weights for the middle 40% of weights are 1109 and 1199.
Normal Probability Distribution
In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- It measures how many standard deviations the measure is from the mean.
- After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
In this problem:
- The mean is of [tex]\mu = 1154[/tex].
- The standard deviation is of [tex]\sigma = 86[/tex].
Item a:
The cutoff for the highest 10% of the weights is the 100 - 10 = 90th percentile, which is X when Z has a p-value of 0.9, so X when Z = 1.28, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.28 = \frac{X - 1154}{86}[/tex]
[tex]X - 1154 = 1.28(86)[/tex]
[tex]X = 1264[/tex]
The cutoff for the highest 10% of pounds is of 1264.
Item b:
The cutoff for the lowest 20% of weights is the 20th percentile, which is X when Z = -0.84, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-0.84 = \frac{X - 1154}{86}[/tex]
[tex]X - 1154 = -0.84(86)[/tex]
[tex]X = 1082[/tex]
The cutoff for the lowest 20% of weights is of 1082.
Item c:
The middle 40% is between the 30th percentile and the 70th percentile, due to the symmetry of the normal distribution, hence:
30th percentile:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-0.525 = \frac{X - 1154}{86}[/tex]
[tex]X - 1154 = -0.525(86)[/tex]
[tex]X = 1109[/tex]
70th percentile:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]0.525 = \frac{X - 1154}{86}[/tex]
[tex]X - 1154 = 0.525(86)[/tex]
[tex]X = 1199[/tex]
The cutoff weights for the middle 40% of weights are 1109 and 1199.
More can be learned about the normal distribution at https://brainly.com/question/24663213
