# A consumer group has determined that the distribution of life spans for gas ranges (stoves) has a mean of 15.0 years and a standard deviation of 4.2 years. The distribution of life spans for electric ranges has a mean of 13.4 years and a standard deviation of 3.7 years. Both distributions are moderately skewed to the right. Suppose we take a simple random sample of 35 gas ranges and a second SRS of 40 electric ranges. Which of the following best describes the sampling distribution of the differe

###### Question:

a. Mean = 1.6 years, standard deviation = 7.9 years, shape: moderately right-skewed.

b. Mean = 1.6 years, standard deviation - 0.92 years, shape: approximately Normal.

c. Mean= 1.6 years, standard deviation = 0.92 years, shape: moderately right skewed.

d. Mean =1.6 years, standard deviation =0.40 years, shape: approximately Normal.

e. Mean =1.6 years, standard deviation =0.40 years, shape: moderately right skewed.

## Answers

Answer:b. Mean = 1.6 years, standard deviation - 0.92 years, shape: approximately Normal. Step-by-step explanation:Central Limit TheoremThe Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.Subtraction of normal Variables:When we subtract normal variables, the mean is the subtraction of the means, while the standard deviation is the square root of the sum of the variances.A consumer group has determined that the distribution of life spans for gas ranges (stoves) has a mean of 15.0 years and a standard deviation of 4.2 years. Sample of 35:This means that:[tex]\mu_G = 15[/tex][tex]s_G = \frac{4.2}{\sqrt{35}} = 0.71[/tex]The distribution of life spans for electric ranges has a mean of 13.4 years and a standard deviation of 3.7 years. Sample of 40:This means that:[tex]\mu_E = 13.4[/tex][tex]s_E = \frac{3.7}{\sqrt{40}} = 0.585[/tex]Which of the following best describes the sampling distribution of the difference in mean life span of gas ranges and electric ranges? Shape is approximately normal.Mean:[tex]\mu = \mu_G - \mu_E = 15 - 13.4 = 1.6[/tex]Standard deviation:[tex]s = \sqrt{s_G^2+s_E^2} = \sqrt{0.71^2+0.585^2} = 0.92[/tex]So the correct answer is given by option b.