The mean weight of pears grown in a certain orchard is 6 ounces with a standard deviation of .5 ounces. A random sample of 30 pears is selected. What is the approximate probability that the sample mean weight is less than 5.5 ounces?

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To solve this problem, we need to use the Central Limit Theorem since we have a sample size greater than 30.

The Central Limit Theorem states that for a random sample of size n, the distribution of the sample mean will be approximately normal with a mean equal to the population mean and a standard deviation equal to the population standard deviation divided by the square root of the sample size.

In this case, the population mean is 6 ounces and the population standard deviation is 0.5 ounces. The sample size is 30.

First, we need to calculate the standard deviation of the sample mean:

Standard deviation of the sample mean = population standard deviation / √ (sample size)
= 0.5 / √ (30)
= 0.5 / 5.477
≈ 0.091

Now, we can calculate the z-score, which measures the number of standard deviations the sample mean is away from the population mean:

z = (sample mean - population mean) / standard deviation of the sample mean
= (5.5 - 6) / 0.091
≈ -5.495

Next, we need to find the probability of obtaining a z-score less than -5.495. We can use a standard normal distribution table or a calculator to find this probability.

Using a standard normal distribution table, we find that the probability of obtaining a z-score less than -5.495 is essentially 0.

Therefore, the approximate probability that the sample mean weight is less than 5.5 ounces is approximately 0.

To find the approximate probability that the sample mean weight is less than 5.5 ounces, we can use the Central Limit Theorem. The Central Limit Theorem states that the distribution of sample means will be approximately normal, regardless of the shape of the original population, as long as the sample size is large enough.

In this case, we have a sample size of 30, which is considered large enough for the Central Limit Theorem to apply.

To calculate the probability, we need to standardize the sample mean using a z-score. The formula for the z-score is:

z = (x - μ) / (σ / √n)

Where:
- z is the z-score
- x is the sample mean weight (5.5 ounces in this case)
- μ is the population mean weight (6 ounces)
- σ is the population standard deviation (0.5 ounces)
- n is the sample size (30)

Plugging in the values, we have:

z = (5.5 - 6) / (0.5 / √30)
= -0.5 / (0.5 / √30)
= -0.5 / (0.5 / 5.477)
= -0.5 / 0.0913
≈ -5.473

Next, we need to find the probability associated with this z-score. We can consult a standard normal distribution table or use a statistical calculator to find this probability.

Looking up a z-score of approximately -5.473 in a standard normal distribution table, we find that the corresponding probability is almost 0.

Therefore, the approximate probability that the sample mean weight is less than 5.5 ounces is very close to 0.

Z = (mean1 - mean2)/standard error (SE) of difference between means

SEdiff = √(SEmean1^2 + SEmean2^2)

SEm = SD/√(n-1)

Since only one SD is provided, you can use just that to determine SEdiff.

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion corresponding to that Z score.