The mean per capita consumption of milk per year is 131 liters with a variance of 841. If a sample of 132 people is randomly selected, what is the probability that the sample mean would be less than 133.5 liters

Respuesta :

Answer:

0.8389 = 83.89% probability that the sample mean would be less than 133.5 liters.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set 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]

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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

The mean per capita consumption of milk per year is 131 liters with a variance of 841.

This means that [tex]\mu = 131, \sigma = \sqrt{841} = 29[/tex]

Sample of 132 people

This means that [tex]n = 132, s = \frac{29}{\sqrt{132}}[/tex]

What is the probability that the sample mean would be less than 133.5 liters?

This is the p-value of Z when X = 133.5. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{133.5 - 131}{\frac{29}{\sqrt{132}}}[/tex]

[tex]Z = 0.99[/tex]

[tex]Z = 0.99[/tex] has a p-value of 0.8389

0.8389 = 83.89% probability that the sample mean would be less than 133.5 liters.

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