Question

Consider two populations for which μ1 = 35, σ1 = 3, μ2 = 20, and σ2 = 4. Suppose that two independent random samples of sizes n1 = 47 and n2 = 54 are selected. Describe the approximate sampling distribution of x1 − x2 (center, spread, and shape). What is the shape of the distribution? The distribution would be non-normal. The distribution is approximately normal. The shape cannot be determined. What is the mean of the distribution? What is the standard deviation of the distribution?

Answer 1

Answer:

The distribution is approximately normal.

The mean and standard deviation are 15 and 0.98 respectively.

Step-by-step explanation:

The population of the random variables X₁ and X₂ are distributed as follows:

$X_{1}\sim (\mu_{1}=35, \sigma_{1}^{2}=3^{2})\\\\X_{2}\sim (\mu_{2}=20, \sigma_{2}^{2}=4^{2})$

Two independent random samples of sizes,

$n_{1}=47\\\\n_{2}=54$

are selected form the two populations.

According to the Central Limit Theorem if we have an unknown population with mean μ and standard deviation σ and appropriately huge random samples (n > 30) are selected from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.  

Then, the mean of the sample means is given by,

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

And the standard deviation of the sample means is given by,

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

Both the sample selected from the two populations are quite large, i.e. $n_{1}=47>30\\\\n_{2}=54>30$

So, according to the central limit theorem the sampling distribution of sample means $\bar X_{1}\ \text{and}\ \bar X_{2}$ can be approximated by the Normal distribution.

Then, the distribution of $\bar X_{1}$ is:

$\bar X_{1}\sim N(35,\ 0.44)$

And the distribution of $\bar X_{2}$ is:

$\bar X_{2}\sim N(20,\ 0.54)$

If two random variables are normally distributed then their linear function is also normally distributed.

So, the distribution of $\bar X_{1}\ - \bar X_{2}$ is Normal and the shape of the distribution is bell-shaped.

The mean of the distribution of $\bar X_{1}\ - \bar X_{2}$ is:

$E(\bar X_{1}\ -\ \bar X_{2})=E(\bar X_{1})-E(\bar X_{2})\\=35-20\\=15$

The standard deviation of the distribution of $\bar X_{1}\ - \bar X_{2}$ is:

$SD(\bar X_{1}-\bar X_{2})=SD(\bar X_{1})+SD(\bar X_{2})\\=0.44+0.54\\=0.98$

*X₁ and X₂ are independent.

Thus, the mean and standard deviation are 15 and 0.98 respectively.