The mean salary per day of all the 5000 employees who work in a company is 150 Rs. and the standard deviation is 15 Rs. Let be x bar is the mean salary per day for a random sample of certain employees selected from this company. Find the mean and standard deviation for sample size 49.
find out the sapling distribution.
To find the mean and standard deviation of a sample, we can use the following formulas:
1. Mean of the Sample (x̄):
The mean of the sample will be the same as the population mean. In this case, it is given that the mean salary per day for all the employees is 150 Rs. Thus, the mean of the sample will also be 150 Rs.
2. Standard Deviation of the Sample (s):
To find the standard deviation of the sample, we need to use the population standard deviation (σ) and the sample size (n). The formula is as follows:
s = σ / √n
Given that the population standard deviation (σ) is 15 Rs and the sample size (n) is 49, we can substitute these values into the formula:
s = 15 / √49
s = 15 / 7
s ≈ 2.14 Rs (rounded to 2 decimal places)
Therefore, the mean of the sample is 150 Rs and the standard deviation of the sample is approximately 2.14 Rs.
As for the sampling distribution, assuming the sample is random and the sample size is large enough (n > 30), we can consider the sampling distribution of the sample mean to be approximately normally distributed. This is known as the Central Limit Theorem.