Data on the oxide thickness of semiconductor wafers are as follows: 425, 431, 416, 419, 421, 436, 418, 410, 431, 433, 423, 426,
DanielleElmas [232]
Answer:
423
Step-by-step explanation:
The sample mean is a point estimate of the population mean.
Therefore a point estimate of the mean oxide thickness for all wafers in the population is the mean of the sample data on the oxide thickness of semiconductor wafers.
To calculate the sample mean, we sum all the sample data and divide by the sample size which is 24. We get 423.33 ≈423
Answer:
2/7 or 0.2857
Step-by-step explanation:
The expected time before the first bulb burns out (two bulbs working) is given by the inverse of the probability that a bulb will go out each day:

The expected time before the second bulb burns out (one bulb working), after the first bulb goes out, is given by the inverse of the probability that the second bulb will go out each day:

Therefore, the long-run fraction of time that there is exactly one bulb working is:

There is exactly one bulb working 2/7 or 0.2857 of the time.
Answer:
r = 0.9825; good correlation.
Step-by-step explanation:
One formula for the correlation coefficient is
![r = \dfrac{n\sum{xy} - \sum{x} \sum{y}}{\sqrt{n\left [\sum{x}^{2}-\left (\sum{x}\right )^{2}\right]\left [\sum{y}^{2}-\left (\sum{y}\right )^{2}\right]}}](https://tex.z-dn.net/?f=r%20%3D%20%5Cdfrac%7Bn%5Csum%7Bxy%7D%20-%20%5Csum%7Bx%7D%20%5Csum%7By%7D%7D%7B%5Csqrt%7Bn%5Cleft%20%5B%5Csum%7Bx%7D%5E%7B2%7D-%5Cleft%20%28%5Csum%7Bx%7D%5Cright%20%29%5E%7B2%7D%5Cright%5D%5Cleft%20%5B%5Csum%7By%7D%5E%7B2%7D-%5Cleft%20%28%5Csum%7By%7D%5Cright%20%29%5E%7B2%7D%5Cright%5D%7D%7D)
The calculation is not difficult, but it is tedious.
1. Calculate the intermediate numbers
We can display them in a table.
<u> </u><u>x</u> <u> y </u> <u> xy </u> <u> x² </u> <u> y² </u>
-3 -40 120 9 1600
1 12 12 1 144
5 72 360 25 5184
<u> 7</u> <u>137</u> <u> 959</u> <u>49</u> <u>18769
</u>
Σ = 10 181 1451 84 25697
2. Calculate the correlation coefficient
![r = \dfrac{n\sum{xy} - \sum{x} \sum{y}}{\sqrt{\left [n\sum{x}^{2}-\left (\sum{x}\right )^{2}\right]\left [n\sum{y}^{2}-\left (\sum{y}\right )^{2}\right]}}\\\\= \dfrac{4\times 1451 - 10\times 181}{\sqrt{[4\times 84 - 10^{2}][4\times25697 - 181^{2}]}}\\\\= \dfrac{5804 - 1810}{\sqrt{[336 - 100][102788 - 32761]}}\\\\= \dfrac{3994}{\sqrt{236\times70027}}\\\\= \dfrac{3994}{\sqrt{16526372}}\\\\= \dfrac{3994}{4065}\\\\= \mathbf{0.9825}](https://tex.z-dn.net/?f=r%20%3D%20%5Cdfrac%7Bn%5Csum%7Bxy%7D%20-%20%5Csum%7Bx%7D%20%5Csum%7By%7D%7D%7B%5Csqrt%7B%5Cleft%20%5Bn%5Csum%7Bx%7D%5E%7B2%7D-%5Cleft%20%28%5Csum%7Bx%7D%5Cright%20%29%5E%7B2%7D%5Cright%5D%5Cleft%20%5Bn%5Csum%7By%7D%5E%7B2%7D-%5Cleft%20%28%5Csum%7By%7D%5Cright%20%29%5E%7B2%7D%5Cright%5D%7D%7D%5C%5C%5C%5C%3D%20%5Cdfrac%7B4%5Ctimes%201451%20-%2010%5Ctimes%20181%7D%7B%5Csqrt%7B%5B4%5Ctimes%2084%20-%2010%5E%7B2%7D%5D%5B4%5Ctimes25697%20-%20181%5E%7B2%7D%5D%7D%7D%5C%5C%5C%5C%3D%20%5Cdfrac%7B5804%20-%201810%7D%7B%5Csqrt%7B%5B336%20-%20100%5D%5B102788%20-%2032761%5D%7D%7D%5C%5C%5C%5C%3D%20%5Cdfrac%7B3994%7D%7B%5Csqrt%7B236%5Ctimes70027%7D%7D%5C%5C%5C%5C%3D%20%5Cdfrac%7B3994%7D%7B%5Csqrt%7B16526372%7D%7D%5C%5C%5C%5C%3D%20%5Cdfrac%7B3994%7D%7B4065%7D%5C%5C%5C%5C%3D%20%5Cmathbf%7B0.9825%7D)
The closer the value of r is to +1 or -1, the better the correlation is. The values of x and y are highly correlated.