Question

A variable of two population has a mean of $7.9$and standard deviation of $5.4$for one of the population and a mean of $7.1$and a standard deviation of $4.6$ for the other population.

a. For independent samples of sizes $3and6$respectively find the mean and standard deviation of $\overline{x_1}-\overline{x_2}$

b. Must the variable under consideration be normally distributed on each of the two population for you to answer part (a) ? Explain your answer.

b. Can you conclude that the variable $\overline{x_1}-\overline{x_2}$is normally distributed? Explain your answer.

Short answer

(a) ${\mu }_1-{\mu }_2=0.8$and ${\sigma }_{\overline{x_1}-\overline{x_2}}=3.64$

(b) No, we cannot conclude that the variable under consideration be normally distributed .

(c) No ,we cannot conclude that the variable is normally distributed.

Step-by-step solution

Part (a)  Step 1: Given Information 

Given in the question that,

Population 1: Mean is $7.9$, standard deviation is $5.4$, and sample size is $3$

Population 2: Mean is $7.1$and standard deviation is $4.6$and sample size is $6$we have to calculate the mean and standard deviation of $\overline{x_1}-\overline{x_2}$

Part (a) Step 2: Explanation 

Mean:

$$\begin{gathered} \overline{x_1}-\overline{x_2}={\mu }_1-{\mu }_2 \\ =7.9-7.1 \\ =0.8 \end{gathered}$$

Standard deviation:

${\sigma }_{{\overline{x}}_1-{\overline{x}}_2}=\sqrt{\frac{{\sigma }_1^2}{n_1}+\frac{{\sigma }_2^2}{n_2}}$

$\Rightarrow {\sigma }_{{\overline{x}}_1-{\overline{x}}_2}=\sqrt{\frac{5.4^2}{3}+\frac{4.6^2}{6}}$

$\Rightarrow {\sigma }_{{\overline{x}}_1-{\overline{x}}_2}=3.64$

hence, mean$=0.8$standard deviation$=3.64$

Part (b)  Step 1: Given Information 

Given in the question that,

Population 1: Mean is $7.9$, standard deviation is $5.4$, and sample size is $3$Population 2: Mean is $7.1$and standard deviation is $4.6$and sample size is

$6$.we have To calculate If the variable under consideration must be normally distributed on each of the population given.

Part (b) Step 2: Explanation 

from part ( a) mean and standard deviation are $0.8and3.64$

Regardless of the distribution of variables on an individual population, the distribution of mean and standard deviation of$\overline{x_1}-\overline{x_2}$remains the same.

The standard deviation formula for $\overline{x_1}-\overline{x_2}$also based on the assumption that the samples are independent.

As a result, the variable under examination does not have to be regularly distributed throughout the entire population.

Part (c)  Step 1: Given Information 

Given in the question that,

Population 1: Mean is $7.9$, standard deviation is $5.4$, and sample size is $3$Population 2: Mean is $7.1$and standard deviation is $4.6$and sample size is$6$

Part (c) Step 2: Explanation 

It is unclear whether the two populations are typically distributed based on the information provided. The sample sizes of$3and6$ are also somewhat tiny. As a result, the variable$\overline{x_1}-\overline{x_2}$ cannot be considered to be regularly distributed.