Mathematics, 26.03.2020 20:08 mikki93

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)

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Answer from: gokusupersaiyan12345

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 $\sigma_{\= Y} = 0.645$

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

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

Answer from: Quest

01010239949494959959699669066060

Answer from: Quest

3, 12, 48, 384, 1536, 6144, 24576, 98304

step-by-step explanation:

just keep timing it by 4.

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