# Essay On Central Limit Theorem Formula

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 mean of the population.
- Standard deviation of the sample is equal to standard deviation of the population divided by square root of 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:

\[\LARGE \mu _{\overline{x}}=\mu\]

$and$

\[\LARGE \sigma _{\overline{x}}=\frac{\sigma }{\sqrt{n}}\]

Where,

μ = Population mean

σ = Population standard deviation

$\mu _{\overline{x}}$ = Sample mean

$\sigma _{\overline{x}}$ = Sample standard deviation

n = Sample size

### Solved Examples

**The record of weights of male population follows normal distribution. Its mean and standard deviation 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?**

**Question 1:**

**Solution:**

Standard deviation of the population = 15 kg

sample size n = 50

Mean of the sample is given by:

$\mu _{\overline{x}}$ = 70 kg

Standard deviation of the sample is given by:

$\sigma _{\overline{x}}$ = $\frac{\sigma }{\sqrt{n}}$

$\sigma _{\overline{x}}$ = $\frac{15}{\sqrt{50}}$

$\sigma _{\overline{x}}$ = 2.121 = 2.1 kg (approx)

Central Limit Theorem states that if we have mean and standard deviation of a particular population and we take a large sample size within the population, then

• Mean of sample is same as mean of the population.

• Standard deviation of the sample is equal to standard deviation of the population divided by square root of sample size.

Central limit theorem is applicable for a sufficiently large sample sizes $(n\ \geq\ 30)$. The formula for central limit theorem can be stated as follows:

**and**

Where,

$\mu$ = Population mean

$s$ = Population standard deviation

$\mu _{\bar{x}}$ = Sample mean

$\sigma _{\bar{x}}$ = Sample standard deviation

$n$ = Sample size.

## Central Limit Theorem Problems

Back to TopFew problems based on central limit theorem are as follows:

### Solved Examples

**Question 1:**The record of weights of male population follows normal distribution. Its mean and standard deviation 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:**

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:

$\mu _{\bar{x}}=\mu$

$\mu _{\bar{x}}$ = 70 kg

Standard deviation of the sample is given by:

$ \sigma _{\bar{x}}$ = $\frac{\sigma }{\sqrt{n}}$

$\sigma _{\bar{x}}$ = $\frac{15}{\sqrt{50}}$

$\sigma _{\bar{x}}$ = 2.121 = 2.1 kg (approx)

**Question 2:**At a coastal area, the number of crabs caught per day are recorded. The average of which is 10 and standard deviation is 3. If the record of 60 days is chosen randomly, estimate the mean and standard deviation of the chosen sample.

**Solution:**

Mean of the population μ = 10

Standard deviation of the population = 3

sample size n = 60

Mean of the sample is given by:

$\mu _{\bar{x}}=\mu$

$\mu _{\bar{x}}$ = 10

Standard deviation of the sample is given by:

$ \sigma _{\bar{x}}$ = $\frac{\sigma }{\sqrt{n}}$

$\sigma _{\bar{x}}$ = $\frac{3}{\sqrt{60}}$

$\sigma _{\bar{x}}$ = 0.387

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