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The weights of bushels of corn produced by a large farm are approximately normally distributed with a mean of 89.46 pounds and a standard deviation of 2.31 pounds. A store is buys 5 bushels of corn from the farm. Assume the 5 bushels the store receives is a random sample of all bushels of corn the farm produces. What is the approximate probability that the 5 bushels will have a mean weight of more than 90 pounds

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MrRoyal

The approximate probability that the 5 bushels will have a mean weight of more than 90 pounds is 0.41

How to determine the probability?

The given parameters are:

Mean, μ = 89.46

Standard deviation, σ = 2.31

The z-score at x = 90 is calculated using

[tex]z = \frac{x - \mu}{\sigma}[/tex]

This gives

[tex]z = \frac{90 - 89.46}{2.31}\\[/tex]

Evaluate

z = 0.234

The probability is then calculated as:

P(x >90) = P(z > 0.234)

From z table of probabilities, we have:

P(x >90) = 0.41

Hence, the approximate probability that the 5 bushels will have a mean weight of more than 90 pounds is 0.41

Read more about probability at:

https://brainly.com/question/251701

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