Respuesta :
Answer:
680kg
Step-by-step explanation:
The average weight of ten bulls is 500kg and the standard deviation of the weight is 30kg. What would be the weight of a bull that is 6 standard deviations above the mean weight
The formula to solve for the weight that is 6 standard deviation above the mean weight =
μ + 6σ = w
Where
μ = Mean weight = 500kg
σ = Standard deviation = 30kg
Hence:
w = 500kg + 6(30kg)
w = 500kg + 180kg
w = 680kg
Therefore, the weight of a bull that is 6 standard deviations above the mean weight is 680 kg
Using the normal distribution, it is found that the weight of a bull that is 6 standard deviations above the mean weight is of 680 kg.
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 and the standard deviation are, respectively, of [tex]\mu = 500[/tex] and of [tex]\sigma = 30[/tex].
The weight of a bull that is 6 standard deviations above the mean weight is X when Z = 6, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]6 = \frac{X - 500}{30}[/tex]
[tex]X - 500 = 6 \times 30[/tex]
[tex]X = 680[/tex]
The weight of a bull that is 6 standard deviations above the mean weight is of 680 kg.
More can be learned about the normal distribution at https://brainly.com/question/24663213
