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Answer :
To find the probability that the mean of the sample will be less than 200 pounds, we will use the concept of the sampling distribution of the sample mean.
First, we need to understand the parameters:
- Population Mean ([tex]\mu[/tex]): 196 pounds
- Population Standard Deviation ([tex]\sigma[/tex]): 22 pounds
- Sample Size ([tex]n[/tex]): 50 individuals
The sample mean ([tex]\bar{x}[/tex]) will approximately follow a normal distribution because the sample size is large ([tex]n \geq 30[/tex]), according to the Central Limit Theorem. The mean of the sample means is the same as the population mean, [tex]\mu = 196[/tex] pounds.
The standard deviation of the sample mean, also known as the standard error (SE), is calculated using the formula:
[tex]SE = \frac{\sigma}{\sqrt{n}}[/tex]
Substituting the given values:
[tex]SE = \frac{22}{\sqrt{50}} \approx 3.11[/tex]
Now, we want to find the probability that the sample mean is less than 200 pounds. We calculate the z-score using the formula:
[tex]Z = \frac{\bar{x} - \mu}{SE}[/tex]
Substitute the given values:
[tex]Z = \frac{200 - 196}{3.11} \approx 1.29[/tex]
Using a standard normal distribution table, or a calculator, we find the probability corresponding to [tex]Z = 1.29[/tex].
The probability that [tex]Z < 1.29[/tex] is approximately 0.9015. Therefore, the probability that the sample mean will be less than 200 pounds is approximately 90.15%.
This means that if we were to take many samples of 50 individuals, we would expect about 90.15% of those samples to have a mean consumption of less than 200 pounds of red meat per year.
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