Mathematics, 12.03.2020 02:26 backyardmastrs

Data on oxide thickness of semiconductors are as follows: 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) Calculate a point estimate of the mean oxide thickness for all wafers in the population. (Round your answer to three decimal places.)(b) Calculate a point estimate of the standard deviation of oxide thickness for all wafers in the population. (Round your answer to two decimal places.)(c) Calculate the standard error of the point estimate from part (a). (Round your answer to two decimal places.)(d) Calculate a point estimate of the median oxide thickness for all wafers in the population. (Express your answer to one decimal places.)(e) Calculate a point estimate of the proportion of wafers in the population that have oxide thickness greater than 430 angstrom. (Round your answer to four decimal places.)

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Answer from: nerdywolf2003

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

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

Answer from: Quest

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Answer from: Quest

a. turk and mcalister

b. howard, mission, and market

c. yes. since streets end some do not cross yet are not parallel to others. example: mcalister and 5th

step-by-step explanation:

parallel lines are lines which do not intersect ever and look a lot like railroad tracks. turk, mcalister and golden gate are examples of these.

intersect are streets which cross each other and a car on one can reach the other. howard, mission, and market intersect with 6th.

skew lines are lines which neither intersect nor are parallel. these lines exist here because some streets end and never intersect another like mcalister and 5th.

A. name the streets that are parallel to golden gate. b. name the streets that intersect 6th street.

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Data on oxide thickness of semiconductors are as follows: 426, 433, 415, 420, 420, 438, 417, 410, 43...