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3 years ago

Question 9 of 10

Mathematics

1 answer:

Lady_Fox [76]3 years ago

3 0

Answer:

the answer is b B is the answer

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Step-by-step explanation:

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3 years ago

Data on oxide thickness of semiconductors are as follows: 426, 433, 415, 420, 420, 438, 417, 410, 430, 434, 423, 426, 412, 434,

Luda [366]

Answer:

a) 423.458

b) 9.53

c) 1.94

d) 424.5

e) 0.2916

Step-by-step explanation:

We are given the following in the question:

426, 433, 415, 420, 420, 438, 417, 410, 430, 434, 423, 426, 412, 434, 435, 432, 409, 426, 409, 436, 422, 430, 411, 415

a) point estimate of the mean oxide thickness

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{10163}{24} = 423.458$

b) point estimate of the standard deviation of oxide thickness

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.

Sum of squares of differences = 2089.958

$s = \sqrt{\dfrac{2089.958}{23}} = 9.53$

c) standard error of the point estimate

Standard error =

$=\dfrac{\sigma}{\sqrt{n}} = \dfrac{9.53}{\sqrt{24}} = 1.94$

d) point estimate of the median oxide thickness

Sorted data: 409, 409, 410, 411, 412, 415, 415, 417, 420, 420, 422, 423, 426, 426, 426, 430, 430, 432, 433, 434, 434, 435, 436, 438

$Median:\\\text{If n is odd, then}\\\\Median = \displaystyle\frac{n+1}{2}th ~term \\\\\text{If n is even, then}\\\\Median = \displaystyle\frac{\frac{n}{2}th~term + (\frac{n}{2}+1)th~term}{2}$

Median =

$\dfrac{12^{th}+13^{th}}{2} = \dfrac{423+426}{2} = 424.5$

e) proportion of wafers in the population that have oxide thickness greater than 430 angstrom

Sample size, n = 24

Number of wafers with oxides greater than 430 angstrom, x = 7

$p = \dfrac{x}{n} = \dfrac{7}{24} = 0.2916$

7 0

4 years ago