Suppose babies born in a large Hospital have a mean weight of 4088 in a variance of 5569. if 128 babies are sample that a random from the hospital what is the probability that the mean weight of the sample babies would differ from the population mean by less than 41 grams round your answer to four decimal places?

To solve this problem, we can use the Central Limit Theorem, which states that for a large sample size, the distribution of the sample mean approximates a normal distribution regardless of the shape of the population distribution.

First, we find the standard deviation of the sample mean. The formula for the standard deviation of the sample mean is given by:

Standard deviation of the sample mean = sqrt(variance of population) / sqrt(sample size)

In this case, the variance of the population is 5569 and the sample size is 128.

Standard deviation of the sample mean = sqrt(5569) / sqrt(128) = 23.578

Next, we calculate the z-score, which represents the number of standard deviations the sample mean is from the population mean. The formula for z-score is given by:

z = (sample mean - population mean) / standard deviation of the sample mean

In this case, we want to find the probability that the mean weight of the sample babies differs from the population mean by less than 41 grams. So the sample mean would be (4088 + 41) = 4129.

z = (4129 - 4088) / 23.578 = 1.743

Now, we can use a standard normal distribution table or a calculator to find the probability associated with this z-score. The probability will represent the area under the normal curve for a z-score of 1.743.

Looking up the z-score of 1.743 in the table or using a calculator, we find that the probability associated with this z-score is approximately 0.9582.

Therefore, the probability that the mean weight of the sample babies would differ from the population mean by less than 41 grams is approximately 0.9582 (rounded to four decimal places).

To solve this problem, we will use the Central Limit Theorem, which states that as the sample size increases, the distribution of the sample means will approach a normal distribution, regardless of the shape of the population distribution.

The mean weight of the population is given as 4088 grams, and the variance is given as 5569 grams. The standard deviation of the population can be calculated by taking the square root of the variance: √5569 ≈ 74.6 grams.

The sample size is 128, and we want to find the probability that the mean weight of the sample babies differs from the population mean by less than 41 grams. In other words, we need to find the probability that the sample mean falls within the range of 4088 ± 41 grams.

To apply the Central Limit Theorem, we need to calculate the standard deviation of the sample mean (also known as the standard error) using the population standard deviation and the sample size. The standard error is given by the formula: standard deviation / √sample size.

Standard error = 74.6 / √128 ≈ 6.59 grams.

Now we have transformed the problem into finding the probability that the sample mean falls within the range of 4088 ± 41 grams, which is equivalent to finding the probability that a standard normal variable (Z) falls within the range of -41 / 6.59 to 41 / 6.59.

Z1 = -41 / 6.59 ≈ -6.22
Z2 = 41 / 6.59 ≈ 6.22

Now, we can use a standard normal distribution table or a calculator to find the probability that Z falls between -6.22 and 6.22. This probability represents the probability that the mean weight of the sample babies differs from the population mean by less than 41 grams.

Using a standard normal distribution table, we find that the probability is practically 1 (or 100%) because the Z-values are extremely large. However, for a more precise answer, you can use a calculator or statistical software.

Therefore, the probability that the mean weight of the sample babies differs from the population mean by less than 41 grams is practically 1 or 100%.

If I recall correctly,

sd = √5569 = appr 74.626

so we want between 4088+41 and 4088-41
that is , between 4129 and 4047

Using my favourite webpage for this stuff,
http://davidmlane.com/hyperstat/z_table.html
and entering the above data, I got a probability of .4173

So for 128 that would be 53 babies