The number of items produced by a small factory is recorded

The number of items produced by a small factory is recorded each day for 134 days of production. The values are labeled x₁,...,x₁₃₄. The daily profit coming from daily production of x items is .25x –320. For this reason, management creates second y -list by y_i = .25x_i–320. This list represents profits (if y_i is negative, the day\'s work is at a loss).

Let x^_ be the mean of the x -list,

let y^_ be the mean of the y -list,

let _x be the (sample) standard deviation for the x -list, and

let _y be the (sample) standard deviation for the y -list.

Answer the following questions.

1.Suppose that x^_ = 1620. Do you have sufficient information to determine y^_? If so, what is it?

2.Suppose that y^_ = 27.50 . Do you have sufficient information to determine x^_? If so, what is it?

3.Suppose that _x = 110. Do you have sufficient information to determine _y? If so, what is it?

4.Suppose that _y = 22.50 . Do you have sufficient information to determine _x? If so, what is it?

5.Find numbers a and b such that, for each index i, 1200 < x_i < 1400 if and only if a < y_i < b.

Solution

The number of items produced by a small factory is recorded each day for 134 days of production.

Let x be the number of days of production and denoted by x1,x2,...........,x134.

n = Number of days of procduction = 134

The daily profit coming from daily production of x items is .25x –320.This profit is only for one particular x value.

yi = .25xi–320 This is also profit which is for i^th value.

And given that if yi is negative then work is at a loss.

Let x_ be the mean of the x -list,

let y_ be the mean of the y -list,

let x be the sample standard deviation for the x -list, and

let y be the sample standard deviation for the y -list.

1. If x_ = 1620,

then from the equation yi = 0.25xi - 320

take summation over i=1,2,3,.....,134

yi = 0.25 * xi - 320

divide by n,

yi / n = 0.25 * xi / n - (320) / n

And we know that yi / n = y_ = mean of y.

Similarly , xi / n = x_ = mean of x.

Therefore , y_ = 0.25 * x_ - n * 320 / n.

y_ = 0.25 * x_ - 320. ____________ a)

Here we only need value of x_ which gives us value of y_ so it is sufficient information to determine y_.

Put value of x_ in equation a),

   y_ = 0.25 * 1620 - 320

   = 405 - 320

= 85

So if x_ = 1620 then y_ = 85.

2.Suppose y_ = 27.50,

From equation a) here also we find x_ if y_ is known i.e. 27.50.

y_ = 0.25 * x_ - 320

27.50 = 0.25 *x_ - 320

27.50 + 320 = 0.25 * x_

347.5 = 0.25 * x_

x_ = 347.5 / 0.25

x_ = 1390

That means if y_ = 27.50 then x_ = 1390.

Here also same concept is there given value of y_ which gives us x_ that is there is sufficient information to determine y_ when given value of x_.

3. If x = 110 then ,

yi = 0.25xi - 320 this equation will be,

We know that the result,

² y = 0.25 * ² x _____b)(variance of constant term is always 0)

taking square root of equation b) so we get,

_y = 0.25 * _x ______c)

Here we only need value of _x to calculate _y and we have given that the value of _{x .}

_y = 0.25 * 110 = 27.5

There is sufficient information to determine _y .

4. _y = 22.50 then,

we use for this also equation c),

_y = 0.25 * _x  

22.50 = 0.25 * _x

_x = 90 .

There is sufficient information to determine _x on the basis of _y = 22.50.

5. Find a and b for each index i , 1200 < xi <1400 if and only if a < yi < b.

We have given that the interval 1200 < xi < 1400.

We convert it into yi ,

yi = 0.25*xi - 320

yi + 320 = 0.25*xi

xi = yi / 0.25 + 320 / 0.25

put this in the following interaval ,

1200 < (yi / 0.25) + (320 / 0.25) < 1400

1200 < (yi / 0.25) + 1280 < 1400

1200 - 1280 < (yi / 0.25) < 1400 -1280

-80 < (yi / 0.25) < 120

-80 * 0.25 < yi < 120 * 0.25

-20 < yi < 30

Therefore a = -20 and b = 30.

Hence an answer.