Suppose the mean length of time that a caller is placed on hold when telephoning a customer service center is 23.8 seconds, with standard deviation 4.6 seconds. Find the probability that the mean length of time on hold in a sample of 1,200 calls will be within 0.5 second of the population mean.

0.5 is 0.1087σ from the mean.

Now run that through your standard formulas.

please oobleck help us complete the standard formulas to arrive at the final answer.

To find the probability that the mean length of time on hold in a sample of 1,200 calls will be within 0.5 second of the population mean, we can use the Central Limit Theorem.

The Central Limit Theorem states that regardless of the shape of the population distribution, the distribution of the sample means will become approximately normally distributed as the sample size increases.

In this case, we have a large sample size of 1,200 calls, which allows us to apply the Central Limit Theorem.

First, we calculate the standard error (SE) using the formula:

SE = standard deviation / square root of sample size

SE = 4.6 / √1200 = 4.6 / 34.64 ≈ 0.1328

Next, we need to find the z-score for a sample mean within 0.5 second of the population mean. We can use the formula:

z = (sample mean - population mean) / SE

For a sample mean within 0.5 second of the population mean:

z = (23.8 - 23.8) / 0.1328 = 0

Now, we can find the probability by looking up the z-score in the standard normal distribution table (or using a calculator or software).

Since the z-score is 0, the probability is 0.5 (or 50%).

Therefore, the probability that the mean length of time on hold in a sample of 1,200 calls will be within 0.5 second of the population mean is 0.5 (or 50%).

To find the probability that the mean length of time on hold in a sample of 1,200 calls will be within 0.5 seconds of the population mean, we need to use the central limit theorem.

The central limit theorem states that the sampling distribution of the sample means will approach a normal distribution as the sample size increases, regardless of the shape of the population distribution.

In this case, since the sample size is large (1,200 calls) and the population standard deviation is known, we can use the normal distribution to approximate the sampling distribution.

First, we need to calculate the standard deviation of the sample means, also known as the standard error of the mean. The formula for the standard error of the mean is:

Standard Error = Population Standard Deviation / √(Sample Size)

Substituting the values given in the question:
Standard Error = 4.6 / √(1200)

Next, we calculate the range of values that are within 0.5 seconds of the population mean. We add and subtract 0.5 seconds from the population mean:

Upper Value = Population Mean + 0.5
Lower Value = Population Mean - 0.5

Finally, we can use the normal distribution to find the probability that the sample mean falls within this range. We can do this by standardizing the range and using the z-score formula:

Z = (x - μ) / σ

Where:
Z = z-score
x = sample mean
μ = population mean
σ = standard error of the mean

Once we calculate the z-scores for both the upper and lower values, we can use a standard normal distribution table or a calculator to find the probability associated with those z-scores.

I'll go ahead and compute the probability for you using the given values:
Standard Error = 4.6 / √(1200) ≈ 0.1324

Upper Value = 23.8 + 0.5 = 24.3
Lower Value = 23.8 - 0.5 = 23.3

Z for the upper value:
Z = (24.3 - 23.8) / 0.1324 ≈ 3.779

Z for the lower value:
Z = (23.3 - 23.8) / 0.1324 ≈ -3.779

To find the probability, we can look up the z-scores in a standard normal distribution table or use a calculator. For example, using a standard normal distribution table, the probability associated with a z-score of 3.779 is approximately 0.9998 and the probability associated with a z-score of -3.779 is also approximately 0.9998.

Therefore, the probability that the mean length of time on hold in a sample of 1,200 calls will be within 0.5 second of the population mean is approximately:

0.9998 - 0.9998 = 0.0000

Note: The probability is very close to 0, which means it is highly unlikely that the mean length of time on hold in a sample of 1,200 calls will be within 0.5 second of the population mean.