Central Limit Theorem Practice Problems

The Central Limit Theorem (CLT) explains why the sample mean tends to follow a normal (bell-shaped) distribution, even when the original data is not normally distributed.

Example 1. The male population's weight data follows a normal distribution. It has a mean of 70 kg and a standard deviation of 15 kg. What would the mean and standard deviation of a sample of 50 guys be if a researcher looked at their records?

Given: μ = 70 kg, σ = 15 kg, n = 50

As per the Central Limit Theorem, the sample mean is equal to the  population mean.

Hence, \mu _{\overline{x}}     = μ = 70 kg

Now, \sigma _{\overline{x}}=\frac{\sigma }{\sqrt{n}}     = 15/√50

⇒ \sigma _{\overline{x}}     ≈ 2.1 kg

Example 2. A distribution has a mean of 69 and a standard deviation of 420. Find the mean and standard deviation if a sample of 80 is drawn from the distribution.

Given: μ = 69, σ = 420, n = 80

As per the Central Limit Theorem, the sample mean is equal to the  population mean.

Hence, \mu _{\overline{x}} = μ = 69 

Now, \sigma _{\overline{x}}=\frac{\sigma }{\sqrt{n}}

⇒ \sigma _{\overline{x}}= 420/√80

⇒ \sigma _{\overline{x}} = 46.95 

Example 3. The mean age of people in a colony is 34 years. Suppose the standard deviation is 15 years. The sample size is 50. Find the mean and standard deviation of the sample.

Given: μ = 34, σ = 15, n = 50

As per the Central Limit Theorem, the sample mean is equal to the  population mean.

Hence, \mu _{\overline{x}}     = μ = 34 years

Now, \sigma _{\overline{x}}=\frac{\sigma }{\sqrt{n}}

⇒ \sigma _{\overline{x}}    = 15/√50

⇒ \sigma _{\overline{x}}     = 2.12 years

Example 4. The mean age of cigarette smokers is 35 years. Suppose the standard deviation is 10 years. The sample size is 39. Find the mean and standard deviation of the sample.

Given: μ = 35, σ = 10, n = 39

As per the Central Limit Theorem, the sample mean is equal to the  population mean.

Hence, \mu _{\overline{x}}     = μ = 35 years

Now, \sigma _{\overline{x}}=\frac{\sigma }{\sqrt{n}}     = 10/√39

⇒ \sigma _{\overline{x}}     = 1.601 years

Example 5. The mean time taken to read a newspaper is 8.2 minutes. Suppose the standard deviation is one minute. Take a sample of size 70. Find its mean and standard deviation.

Given: μ = 8.2, σ = 1, n = 70

As per the Central Limit Theorem, the sample mean is equal to the  population mean.

Hence, \mu _{\overline{x}}     = μ = 8.2 minutes

Now, \sigma _{\overline{x}}=\frac{\sigma }{\sqrt{n}}     = 1/√70

⇒ \sigma _{\overline{x}}     = 0.11 minutes

Example 6. A distribution has a mean of 12 and a standard deviation of 3. Find the mean and standard deviation if a sample of 36 is drawn from the distribution.

Given: μ = 12, σ = 3, n = 36

As per the Central Limit Theorem, the sample mean is equal to the  population mean.

Hence, \mu _{\overline{x}}     = μ = 12

Now, \sigma _{\overline{x}}=\frac{\sigma }{\sqrt{n}}     = 3/√36

⇒ \sigma _{\overline{x}}     = 0.5

Example 7. You want to estimate the mean income of a population with a margin of error of $5, assuming the population standard deviation is $50, and you want a 95% confidence level. What sample size do you need?

Given: Z= 1.96 (for 95% confidence level), σ = 50, E = 5

As per the Central Limit Theorem, the formula to calculate the sample size is.

Hence, n = \left( \frac{Z \times \sigma}{E}\right)^2

n = \left( \frac{1.96 \times 50} {5} \right)^2 = \left( \frac{98}{5} \right)^2 = (19.6)^2

n = 384.16 (Round up to the nearest whole number)
n = 385

The required sample size is 385.

Example 8. Given that the population proportion p=0.40p=0.40p=0.40 and the sample size n=100, calculate the standard error for the sample proportion \hat{p}.

Given: n=100, p=40% or .40.

As per the Central Limit Theorem, the formula to calculate standard error for proportions.

\sigma_{\hat{p}} = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{100}{0.40(1 - 0.40)}} = \sqrt{\frac{100}{0.24}} = 0.04899

\sigma_{\hat{p}}=0.04899

\sigma_{\hat{p}} \approx 0.04899

Practice Problem

Question 1. Given that the population mean is 50 and the population standard deviation is 10, find the Z-score for a sample mean of 52 when the sample size is 25.

Question 2. If the population has a standard deviation of 15, and you take a sample of 50 from this population, calculate the standard error of the sample mean.

Question 3. A population has a mean of 100 and a standard deviation of 20. You take a sample of 36. Calculate the 95% confidence interval for the sample mean.

Question 4. The average height of adult women in a population is 160 cm with a standard deviation of 10 cm. What is the probability that a random sample of 25 women has a mean height greater than 162 cm?

Answer:-

1. 1
2. 2.12
3. [93.47, 106.53]
4. 0.1587