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Answer :
Final answer:
The probability that the combined weight of 40 randomly selected one-year-old Hereford bulls exceeds 46,000 pounds is approximately 16.35%. This calculation uses the Z-score related to the sample mean weight exceeding 1,150 pounds.
Explanation:
To find the probability that the combined weight of 40 randomly selected one-year-old Hereford bulls exceeds 46,000 pounds, we first need to determine if the average weight per bull exceeds 1,150 pounds (since 46,000 ÷ 40 = 1,150). The mean weight of a Hereford bull is given as 1,135 pounds with a standard deviation of 97 pounds. Using the Central Limit Theorem, we can find the mean and standard deviation for the sample mean weight of 40 bulls.
The standard deviation of the sample mean (σ_x-bar) is σ ÷ √{n} = 97 pounds ÷ √{40}, which equals approximately 15.34 pounds. To find the Z-score, we use the formula Z = (X - μ) ÷ σ_x-bar, where X is 1,150 pounds, μ (the mean) is 1,135 pounds, and σ_x-bar is 15.34 pounds, yielding a Z-score of about 0.98.
Using the standard normal distribution table, a Z-score of 0.98 corresponds to a probability of approximately 0.8365. However, since we want the probability that the sample mean is greater than 1,150 pounds, we need to look at the upper tail, which is 1 - 0.8365 = 0.1635. Therefore, there's a 16.35% chance the combined weight of 40 randomly selected one-year-old Hereford bulls exceeds 46,000 pounds.
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Answer:
16.35% probability their combined weight exceeds 46000 pounds.
Step-by-step explanation:
To solve this question, we have to understand the normal probability distribution and the central limit theorem.
Normal probability distribution:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Central limit theorem:
The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sample means with size n of at least 30 can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]
In this problem, we have that:
[tex]\mu = 1135, \sigma = 97, n = 40, s = \frac{97}{\sqrt{40}} = 15.34[/tex]
Find the probability their combined weight exceeds 46000 pounds.
This is 1 subtracted by the pvalue of Z when [tex]X = \frac{46400}{40} = 1150[/tex]. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{1150 - 1135}{15.34}[/tex]
[tex]Z = 0.98[/tex]
[tex]Z = 0.98[/tex] has a pvalue of 0.8365
1 - 0.8365 = 0.1635
16.35% probability their combined weight exceeds 46000 pounds.
