The standard deviation of the masses of articles in a large population is 4.55 kg. Random samples of size 100 are drawn from the population. Find the probability that a sample mean will differ from the population mean by less than 0.8 kg.
To find the probability that a sample mean will differ from the population mean by less than 0.8 kg, we can use the formula for the standard error of the mean:
Standard Error = Standard Deviation / √(Sample Size)
Here, the standard deviation of the population is given as 4.55 kg, and the sample size is 100. So, plugging these values into the formula, we get:
Standard Error = 4.55 / √100 = 4.55 / 10 = 0.455 kg
Now, we need to find the probability that the difference between the sample mean and the population mean is less than 0.8 kg. To do this, we can use the concept of the z-score.
The formula for the z-score is:
z = (x - μ) / σ
Where:
x = Sample mean
μ = Population mean
σ = Standard Deviation
In this case, we want to find the probability that the z-score is less than 0.8/0.455.
z = (0.8 - 0) / 0.455 = 1.76
Now, we need to find the probability corresponding to this z-score. You can use a standard normal distribution table, such as the Z-table or a cumulative distribution function (CDF) calculator, to find the probability associated with this z-score.
Looking up the z-score of 1.76 in the standard normal distribution table, we find that the probability is approximately 0.9608.
Therefore, the probability that a sample mean will differ from the population mean by less than 0.8 kg is approximately 0.9608, or 96.08%.