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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)

Respuesta :

okpalawalter8

Answer:

a

i So  the approximate distribution of [tex]\= X[/tex] is [tex]\mu_{\= X} =103[/tex] and  [tex]\sigma_{\= X} = 0.783[/tex]

ii So the approximate distribution of [tex]\= Y[/tex] is [tex]\mu_{\= Y} =105[/tex] and  [tex]\sigma_{\= Y} = 0.645[/tex]

b

 the approximate distribution of  [tex]\=X - \= Y[/tex] is [tex]E (\= X - \= Y) = -2[/tex] and  [tex]\sigma_{\= X - \=Y}=1.029[/tex]

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

c

the value of  [tex]P(-1 \le \=X - \= Y \le 1)[/tex] is [tex]= -0.1639[/tex]    

Step-by-step explanation:

From the question we are given that

       The expected tensile strength of the type A steel is  [tex]\mu_A = 103 ksi[/tex]

        The standard deviation of type A steel is  [tex]\sigma_A = 7ksi[/tex]

         The expected tensile strength of the type B steel is [tex]\mu_B = 105\ ksi[/tex]

            The standard deviation of type B steel is  [tex]\sigma_B = 5 \ ksi[/tex]

Also the assumptions are

       Let [tex]\= X[/tex] be the sample average tensile strength of a random sample of 80 type-A specimens

Here [tex]n_a =80[/tex]

      Let [tex]\= Y[/tex] be  the sample average tensile strength of a random sample of 60 type-B specimens.

  Here [tex]n_b = 60[/tex]

Let the sampling distribution of the mean be

             [tex]\mu _ {\= X} = \mu[/tex]

                   [tex]=103[/tex]

 Let the sampling distribution of the standard deviation be

               [tex]\sigma _{\= X} = \frac{\sigma }{\sqrt{n_a} }[/tex]

                     [tex]= \frac{7}{\sqrt{80} }[/tex]

                    [tex]=0.783[/tex]

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

For [tex]\= Y[/tex]

 The sampling distribution of the sample mean is

               [tex]\mu_{\= Y} = \mu[/tex]

                    [tex]= 105[/tex]

  The sampling distribution of the standard deviation is

               [tex]\sigma _{\= Y} = \frac{\sigma }{\sqrt{n_b} }[/tex]

                    [tex]= \frac{5}{\sqrt{60} }[/tex]

                    [tex]= 0.645[/tex]

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

Now to obtain the approximate distribution for [tex]\=X - \= Y[/tex]

               [tex]E (\= X - \= Y) = E (\= X) - E(\= Y)[/tex]

                                [tex]= \mu_{\= X} - \mu_{\= Y}[/tex]

                                [tex]= 103 -105[/tex]

                                [tex]= -2[/tex]

The standard deviation of [tex]\=X - \= Y[/tex] is

               [tex]\sigma_{\= X - \=Y} = \sqrt{\sigma_{\= X}^2 - \sigma_{\= Y}^2}[/tex]

                         [tex]= \sqrt{(0.783)^2 + (0.645)^2}[/tex]

                         [tex]=1.029[/tex]

Now to find the value of  [tex]P(-1 \le \=X - \= Y \le 1)[/tex]

  Let us assume that [tex]F = \= X - \= Y[/tex]

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

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

                             [tex]= P[0.972 \le Z \le 2.95][/tex]

                             [tex]= P(Z \le 0.972) - P(Z \le 2.95)[/tex]

Using the z-table to obtain their z-score

                             [tex]= 0.8345 - 0.9984[/tex]

                             [tex]= -0.1639[/tex]

                   

(a): The approximate distribution of [tex]\bar{X}[/tex] is [tex]\mu_{\bar{x}}=105[/tex] and [tex]\sigma_{\bar{x}}=1.265[/tex]

The approximate distribution of [tex]\bar{Y}[/tex] is [tex]\mu_{\bar{y}}=100[/tex] and [tex]\sigma_{\bar{y}}=1.0142[/tex]

(b): The approximate distribution of [tex]\bar{X}-\bar{Y}[/tex] is [tex]E(\bar{X}-\bar{Y})=5[/tex] and [tex]\sigma_k=1.6213[/tex]

(c): The value of [tex]P(-1\le K\le1)=0.0066[/tex]

Probability Distribution:

The probability distribution gives the possibility of each outcome of a random experiment or event. It provides the probabilities of different possible occurrences.

Given that,

[tex]For\ type\ A,\mu=105 ksi\\\sigma=8 ksi\\For\ type\ B,\mu=100ksi\\\sigma=6ksi[/tex]

Part(a):

The sampling distribution of sample mean is,

[tex]\mu_{\bar{x}}=\mu\\=105[/tex]

The sampling distribution of standard deviation is,

[tex]\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n} } \\=\frac{8}{\sqrt{40} } \\=1.265[/tex]

The sampling distribution of sample mean is,

[tex]\mu_{\bar{y}}=100[/tex]

The sampling distribution of standard deviation is,

[tex]\sigma_{\bar{y}}=\frac{\sigma}{\sqrt{n} } \\=\frac{6}{\sqrt{35} } \\=1.0142[/tex]

Part(b):

Let [tex]K=\bar{X}-\bar{Y}[/tex]

The mean of [tex]\bar{X}-\bar{Y}[/tex] is,

[tex]E(\bar{X}-\bar{Y})=E(\bar{X})-E(\bar{Y})\\=105-100\\=5[/tex]

The standard deviation of [tex]\bar{X}-\bar{Y}[/tex] is,

[tex]\sigma_k=\sqrt{\sigma_{\bar{x}}^{2}+\sigma_{\bar{y}}^{2}} \\=\sqrt{(1.265)^2+(1.014)^2}\\ =1.6213[/tex]

Par(c):

Let [tex]K=\bar{X}-\bar{Y}[/tex]

[tex]P(-1\le K\le1)=P(\frac{-1-E(K)}{\sigma_k}\le Z \le \frac{1-E(K)}{\sigma_k} )\\=P(-3.7\le Z\le-2.47\\=\phi(-2.47)-\phi(-3.7)\\=0.006756-0.000108\\=0.0066[/tex]

Learn more about the topic Probability Distribution:

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