Question

Suppose the expected tensile strength of type-A steel is 103 ksi and the standard deviation of tensile strength is 7 ksi. For type-B steel, suppose the expected tensile strength and standard deviation of tensile strength are 105 ksi and 5 ksi, respectively. Let X^- = the sample average tensile strength of a random sample of 80 type-A specimens, and let Y^- = the sample average tensile strength of a random sample of 60 type-B specimens.
a. What is the approximate distribution of X^-? Of Y^-?
b. What is the approximate distribution of (X^-) - (Y^-)? Justify your answer.
c. Calculate (approximately) P(-1<_ X^- - Y^- <_1)

Answer 1

Answer:

a

i So  the approximate distribution of $\= X$ is $\mu_{\= X} =103$ and  $\sigma_{\= X} = 0.783$

ii So the approximate distribution of $\= Y$ is $\mu_{\= Y} =105$ and  $\sigma_{\= Y} = 0.645$

b

 the approximate distribution of  $\=X - \= Y$ is $E (\= X - \= Y) = -2$ and  $\sigma_{\= X - \=Y}=1.029$

Here we can see that the mean of the approximate distribution is negative which tell us that this negative value of the  data for  $\=X - \= Y$ sample   are more and their frequency occurrence is higher than the positive values  

c

the value of  $P(-1 \le \=X - \= Y \le 1)$ is $= -0.1639$    

Step-by-step explanation:

From the question we are given that

       The expected tensile strength of the type A steel is  $\mu_A = 103 ksi$

        The standard deviation of type A steel is  $\sigma_A = 7ksi$

         The expected tensile strength of the type B steel is $\mu_B = 105\ ksi$

            The standard deviation of type B steel is  $\sigma_B = 5 \ ksi$

Also the assumptions are

       Let $\= X$ be the sample average tensile strength of a random sample of 80 type-A specimens

Here $n_a =80$

      Let $\= Y$ be  the sample average tensile strength of a random sample of 60 type-B specimens.

  Here $n_b = 60$

Let the sampling distribution of the mean be

             $\mu _ {\= X} = \mu$

                   $=103$

 Let the sampling distribution of the standard deviation be

               $\sigma _{\= X} = \frac{\sigma }{\sqrt{n_a} }$

                     $= \frac{7}{\sqrt{80} }$

                    $=0.783$

So What this mean is that the approximate distribution of $\= X$ is $\mu_{\= X} =103$ and  $\sigma_{\= X} = 0.783$

For $\= Y$

 The sampling distribution of the sample mean is

               $\mu_{\= Y} = \mu$

                    $= 105$

  The sampling distribution of the standard deviation is

               $\sigma _{\= Y} = \frac{\sigma }{\sqrt{n_b} }$

                    $= \frac{5}{\sqrt{60} }$

                    $= 0.645$

So What this mean is that the approximate distribution of $\= Y$ is $\mu_{\= Y} =105$ and  $\sigma_{\= Y} = 0.645$                      

Now to obtain the approximate distribution for $\=X - \= Y$

               $E (\= X - \= Y) = E (\= X) - E(\= Y)$

                                $= \mu_{\= X} - \mu_{\= Y}$

                                $= 103 -105$

                                $= -2$

The standard deviation of $\=X - \= Y$ is

               $\sigma_{\= X - \=Y} = \sqrt{\sigma_{\= X}^2 - \sigma_{\= Y}^2}$

                         $= \sqrt{(0.783)^2 + (0.645)^2}$

                         $=1.029$

Now to find the value of  $P(-1 \le \=X - \= Y \le 1)$

  Let us assume that $F = \= X - \= Y$

    $P(-1 \le F \le 1)$ $= P [\frac{-1 -E (F)}{\sigma_F} \le Z \le \frac{1-E(F)}{\sigma_F} ]$

                             $= P[\frac{-1-(-2)}{1.029} \le Z \le \frac{1-(-2)}{1.029} ]$

                             $= P[0.972 \le Z \le 2.95]$

                             $= P(Z \le 0.972) - P(Z \le 2.95)$

Using the z-table to obtain their z-score

                             $= 0.8345 - 0.9984$

                             $= -0.1639$