Answer:
b. Mean = 1.6 years, standard deviation - 0.92 years, shape: approximately Normal.
Step-by-step explanation:
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean
and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
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:
![\mu_G = 15](https://tex.z-dn.net/?f=%5Cmu_G%20%3D%2015)
![s_G = \frac{4.2}{\sqrt{35}} = 0.71](https://tex.z-dn.net/?f=s_G%20%3D%20%5Cfrac%7B4.2%7D%7B%5Csqrt%7B35%7D%7D%20%3D%200.71)
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:
![\mu_E = 13.4](https://tex.z-dn.net/?f=%5Cmu_E%20%3D%2013.4)
![s_E = \frac{3.7}{\sqrt{40}} = 0.585](https://tex.z-dn.net/?f=s_E%20%3D%20%5Cfrac%7B3.7%7D%7B%5Csqrt%7B40%7D%7D%20%3D%200.585)
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:
![\mu = \mu_G - \mu_E = 15 - 13.4 = 1.6](https://tex.z-dn.net/?f=%5Cmu%20%3D%20%5Cmu_G%20-%20%5Cmu_E%20%3D%2015%20-%2013.4%20%3D%201.6)
Standard deviation:
![s = \sqrt{s_G^2+s_E^2} = \sqrt{0.71^2+0.585^2} = 0.92](https://tex.z-dn.net/?f=s%20%3D%20%5Csqrt%7Bs_G%5E2%2Bs_E%5E2%7D%20%3D%20%5Csqrt%7B0.71%5E2%2B0.585%5E2%7D%20%3D%200.92)
So the correct answer is given by option b.