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Gnesinka [82]
3 years ago
14

Suppose I work on a factory line doing quality control work. Since I've been working here for such a long time, I am fairly conf

ident that 15% of all items (independent of all other items) need to be sent back to be re-worked. These items take a fairly long time to check properly, but it is the same set of steps every time, so I always check 25 items every shift. Let X be the number of items sent back to be re-worked during a given shift. What is the distribution of X? Give the name of the distribution and the appropriate parameters. What is the mean and variance of this distribution? Give your answer as a number, but include the formulas (or logic) used. What is the probability I send back less than 4 items to be re-worked during a single shift? What is the probability I send back exactly 3 items to be re-worked during a single shift?
Mathematics
1 answer:
Dmitriy789 [7]3 years ago
8 0

Answer:

Step-by-step explanation:

a) It is a binomial distribution.

n = 25

p = 0.15

b) mean = np = 25 * 0.15 = 3.75

   variance = np(1 - p)

                 = 25 * 0.15 * (1 - 0.15)

                 = 3.1875

c) P(X = x) = nCx * px * (1 - p)n - x

P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

            = 25C0 * (0.15)^0 * (0.85)^25 + 25C1 * (0.15)^1 * (0.85)^24 + 25C2 * (0.15)^2 * (0.85)^23 + 25C3 * (0.15)^3 * (0.85)^22 = 0.4711

d) P(X = 3) = 25C3 * (0.15)^3 * (0.85)^22 = 0.2174

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But x+y = 15 , then 10x+10y =150 ==> 150=3xy and xy = 50

Now we have the sum S of the 2 parts that is S = 15 and 
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If the equation of the circle is x^2+ y^2 = 41, we must first understand the parts of the equation.
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(h.k) is the radius of the circle
r is the radius of the circle
Another useful fact to know is that tangent lines touch the circle at one point (4,5)

Since in our original equation there are no h or k values, we can assume that the center of the circle is (0,0). 

The formula for slope is <u>Y1-Y2</u>
                                   X1-X2
We can break this down with our two points (center and tangent)
  (0,0)    and   (-4,-5)
(X1,Y1) and (X2,Y2)

therefore, we will put the equation as such
<u>0-(-5)= 5</u>  = <em> </em><u><em>5</em></u>
0-(-4)= 4     <em> 4</em>

<em>this is our slope from the center to the point of tangency.</em>

We know that tangent lines are perpendicular to the radius, which we've already found the slope of. Perpendicular lines are opposite reciprocals of the line they are perpendicular to. 

Therefore, we take our slope from center to the tangent, and make it opposite and then take the reciprocal of that slope, which will give us the slope of the tangent line itself. (note: reciprocal means flip the numerator and denominator)
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</u>4     4        5   

Now, we have a point on the line, and the line's slope. We can use slope-intercept equation to find the equation of the line.

Slope-int      y=mx+b
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b is the y intercept  ( the point where x=0, or where its on the y axis)

now we plug things in
(-4,-5) is our point,
<u>-4</u>  is our slope
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-5=<u>-4</u>(-4)+b       After we plug things in, solve for b
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-5= 3.2+b

-1.8= b  or  b=  <u />1 <u>4</u>
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Now we just need to rewrite our equation with all our components.
(-4.-5) = point
<u>-4</u>  = slope<u>
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1 <u>4</u> = y-intercept<u>
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</u><em>    5         5</em>


Hope that helped
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