A procurement specialist has purchased 25 resistors from vendor 1 and 30 resistors from vendor 2. Let X1,1,X1,2,...,X1,25 represent the vendor 1 observed resistances, which are assumed to be normally and independently distributed with mean 100 ohms and standard deviation 1.5 ohms. Similarly, let X2,1,X2,2,...,X2,30 represent the vendor 2 observed resistances, which are assumed to be normally and independently distributed with mean 105 ohms and standard deviation of 2.0 ohms. What is the sampling distribution of X¯¯¯¯1−X¯¯¯¯2 ? What is the standard error of X¯¯¯¯1−X¯¯¯¯2

Answer:Normal DistributionStandard error = 0.473Step-by-step explanation:We are given the following information:A procurement specialist has purchased 25 resistors from vendor 1  are assumed to be normally distributed with mean 100 ohms and standard deviation 1.5 ohms. [tex]x_{1,1}, x_{1,2}, ... , x_{1,25}\\\mu_{1} = 100\\\sigma_1 = 1.5[/tex]A procurement specialist has purchased 30 resistors from vendor 2 are assumed to be normally distributed with mean 105 ohms and standard deviation 2.0 ohms. [tex]x_{2,1}, x_{2,2}, ... , x_{2,30}\\\mu_{2} = 105\\\sigma_1 = 2.0[/tex]The difference of two independent normally distributed random variables is normal, with its mean being equal to the difference of the two means, and its variance being the sum of the two variances.Thus, the sampling distribution of [tex]X_1-X_2[/tex] is a normal distribution.Mean = [tex]100 - 105 = -5[/tex]Standard error =[tex]\sqrt{\displaystyle\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}\\\\=\sqrt{\displaystyle\frac{(1.5)^2}{25} + \frac{(2)^2}{30}}\\\\= 0.473[/tex]