A farm is sold for €457 000, which gives a profit of 19%. Find the profit

Consider the following sample of observations on coating thickness for low-viscosity paint.

Julli [10]

Answer:

a) $\bar X = \frac{\sum_{i=1}^n X_i}{n}$

And for this case if we use this formula we got:

$\bar x = 1.3538$

b) Since we have n =16 values for the sample the median can be calculated as the average between position 8th anf 9th and we got:

$Median = \frac{1.31+1.46}{2}= 1.385$

c) $P(X>a)=0.1$ (a)

$P(X$ (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.9 of the area on the left and 0.1 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.9 and P(z>1.28)=0.1

If we use condition (b) from previous we have this:

$P(X$

$P(z$

But we know which value of z satisfy the previous equation so then we can do this:

$z=1.28$

And if we solve for a we got

$a=1.3538 +1.28*0.3505=1.8024$

So the value of height that separates the bottom 90% of data from the top 10% is 1.8024.

d) $Median= \frac{x_{8} +x_{9}}{2}$

The variance for this estimator is given by:

$Var(\frac{x_{8} +x_{9}}{2}) = \frac{1}{4} Var(X_{8} +X_{9})$

We can assume the obervations independent so then we have:

$Var(\frac{x_{8} +x_{9}}{2}) = \frac{1}{4} (2\sigma^2) = \frac{\sigma^2}{2}$

And replacing we got:

$Var(\frac{x_{8} +x_{9}}{2})= \frac{0.3105^2}{2}= 0.0482$

And the standard error would be given by:

$Sd(\frac{x_{8} +x_{9}}{2})= \sqrt{0.0482}=0.2196$

Step-by-step explanation:

Data given:

0.86 0.88 0.88 1.07 1.09 1.17 1.29 1.31 1.46 1.49 1.59 1.62 1.65 1.71 1.76 1.83

Part a

We can calculate the mean with the following formula:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

And for this case if we use this formula we got:

$\bar x = 1.3538$

Part b

For this case in order to calculate the median we need to put the data on increasing way like this:

0.86 0.88 0.88 1.07 1.09 1.17 1.29 1.31 1.46 1.49 1.59 1.62 1.65 1.71 1.76 1.83

Since we have n =16 values for the sample the median can be calculated as the average between position 8th anf 9th and we got:

$Median = \frac{1.31+1.46}{2}= 1.385$

Part c

For this case we can assume that the mean is $\mu = 1.3538$

And we can calculate the population deviation with the following formula:

$\sigma = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{N}}$

And if we replace we got: $\sigma= 0.3105$

And assuming normal distribution we have this:

$X \sim N (\mu = 1.3538, \sigma= 0.3105)$

For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.1$ (a)

$P(X$ (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.9 of the area on the left and 0.1 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.9 and P(z>1.28)=0.1

If we use condition (b) from previous we have this:

$P(X$

$P(z$

But we know which value of z satisfy the previous equation so then we can do this:

$z=1.28$

And if we solve for a we got

$a=1.3538 +1.28*0.3505=1.8024$

So the value of height that separates the bottom 90% of data from the top 10% is 1.8024.

Part d

The median is defined as :

$Median= \frac{x_{8} +x_{9}}{2}$

The variance for this estimator is given by:

$Var(\frac{x_{8} +x_{9}}{2}) = \frac{1}{4} Var(X_{8} +X_{9})$

We can assume the obervations independent so then we have:

$Var(\frac{x_{8} +x_{9}}{2}) = \frac{1}{4} (2\sigma^2) = \frac{\sigma^2}{2}$

And replacing we got:

$Var(\frac{x_{8} +x_{9}}{2})= \frac{0.3105^2}{2}= 0.0482$

And the standard error would be given by:

$Sd(\frac{x_{8} +x_{9}}{2})= \sqrt{0.0482}=0.2196$

6 0

2 years ago

From recent polls about customer satisfaction, you know that 85% of the clients of your company are highly satisfied and want to

Lubov Fominskaja [6]

Answer:it doesnt matter what you choose for this one a or b

Step-by-step explanation:

7 0

2 years ago

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What is the selling price of an item if the original cost is $784.50 and the markup on the item is 6.5 percent?

kotegsom [21]

Answer:

835.49

Step-by-step explanation:

selling price = original cost + markup value

We need to find the markup

markup = original cost * markup percent

= $784.50 * 6.5%

= $784.50 *.06.5

=50.9925

Rounding to the nearest cent

=50.99

selling price = original cost + markup value

=784.50+50.99

835.49

6 0

2 years ago

Let A be a 4×4 matrix and suppose that det(A)=3. For each of the following row operations, determine the value of det(B), where

spin [16.1K]

Answer:

a)3

b) 9

c) -3

Step-by-step explanation:

a) If B is obtained by adding a multiple of a row of A to another row of A, $det (B) = det (A).$

Then, $det(B)=3$ .

b) If B is obtained by multiplying a row of A by k, then $det (B) = kdet (A)$ . Then $det(B)=3det(A)=3*3=9$

c) If B is obtained by exchanging two rows (columns) of A, then $det (B) = - det (A)$ . Then $det (B) = - 3$

6 0

2 years ago

Of the 400 eighth-graders at pascal middle school, 117 take algebra, 109 take advanced computer, and 114 take industrial technol

Hatshy [7]

If 164 of the 400 eighth-graders take none of these courses, then 400-164=236 students take some courses.

Let x students take all three courses, then:

1. 70-x take only both algebra and advanced computer,

2. 34-x take only both algebra and industrial technology,

3. 29-x take only both advanced computer and industrial technology.

Now let's count how many student take only one course:

1. 117-(x+34-x+70-x)=13+x take algebra,

2. 109-(x+29-x+70-x)=10+x take advanced computer,

3. 114-(x+29-x+34-x)=51+x take industrial technology.

Now count all students that take some courses:

51+x+10+x+13+x+29-x+34-x+70-x+x=236,

x=236-51-10-13-29-34-70,

x=29.

Answer: all three courses take 29 students.

6 0

2 years ago

Read 2 more answers

A farm is sold for €457 000, which gives a profit of 19%. Find the profit​

A farm is sold for €457 000, which gives a profit of 19%. Find the profit