The Central Limit Theorem is the sampling distribution of the sampling means approaches a normal distribution as the sample size gets larger, no matter what the shape of the data distribution. An essential component of the Central Limit Theorem is the average of sample means will be the population mean.
Similarly, if you find the average of all of the standard deviations in your sample, you will find the actual standard deviation for your population.
- Mean of sample is same as the mean of the population.
- The standard deviation of the sample is equal to the standard deviation of the population divided by the square root of the sample size.
Central limit theorem is applicable for a sufficiently large sample sizes (n ≥ 30). The formula for central limit theorem can be stated as follows:
\[\LARGE \mu _{\overline{x}}=\mu\]
\(\begin{array}{l}and\end{array} \)
\[\LARGE \sigma _{\overline{x}}=\frac{\sigma }{\sqrt{n}}\]
Where,
μ = Population mean
σ = Population standard deviation
\(\begin{array}{l}\mu _{\overline{x}}\end{array} \)
= Sample mean\(\begin{array}{l}\sigma _{\overline{x}}\end{array} \)
= Sample standard deviationn = Sample size
Solved Example
Question:Â The record of weights of the male population follows the normal distribution. Its mean and standard deviations are 70 kg and 15 kg respectively. If a researcher considers the records of 50 males, then what would be the mean and standard deviation of the chosen sample?
Solution:
Solution:
Mean of the population μ = 70 kg
Standard deviation of the population = 15 kg
sample size n = 50
Mean of the sample is given by:
Standard deviation of the sample is given by:
Standard deviation of the population = 15 kg
sample size n = 50
Mean of the sample is given by:
\(\begin{array}{l}\mu _{\overline{x}}\end{array} \)
= 70 kgStandard deviation of the sample is given by:
\(\begin{array}{l}\sigma _{\overline{x}}\end{array} \)
= \(\begin{array}{l}\frac{\sigma }{\sqrt{n}}\end{array} \)
\(\begin{array}{l}\sigma _{\overline{x}}\end{array} \)
= \(\begin{array}{l}\frac{15}{\sqrt{50}}\end{array} \)
\(\begin{array}{l}\sigma _{\overline{x}}\end{array} \)
= 2.122 = 2.1 kg (approx)
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