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nexus9112 [7]
2 years ago
15

A drawer contains 4 capsules numbered 2, 3, 5, and 8. a sample of size 3 is drawn with replacement. what is the standard deviati

on of x-bar?
Mathematics
1 answer:
Snezhnost [94]2 years ago
7 0
The table below summarises the calculation for the standard deviation of x-bar.

\begin{tabular}
{|c|c|c|c|}
Drawing (x)&$\bar{x}$&$x-\bar{x}$&$(x-\bar{x})^2\\[1ex]
2,2,2&2&-2.5&6.25\\
2,2,3&2.3&-2.2&4.84\\
2,2,5&3&-1.5&2.25\\
2,2,8&4&-0.5&0.25\\
2,3,3&2.7&-1.8&3.24\\
2,3,5&3.3&-1.2&1.44\\
2,3,8&4.3&-0.2&0.04\\
2,5,5&4&-0.5&0.25\\
2,5,8&5&0.5&0.25\\
2,8,8&6&1.5&2.25\\
3,3,3&3&-1.5&2.25\\
3,3,5&3.7&-0.8&0.64\\
3,3,8&4.7&0.2&0.04\\
3,5,5&4.3&-0.2&0.04\\
3,5,8&5.3&0.8&0.64\\
3,8,8&6.3&1.8&3.24\\
5,5,5&5&0.5&0.25\\
5,5,8&6&1.5&2.25\\
5,8,8&7&2.5&6.25\\
8,8,8&8&3.5&12.25\\
\end{tabular}

\sum\bar{x}=89.9 \\  \\ \bar{\bar{x}}= \frac{\sum\bar{x}}{n}=\frac{89.9}{20} \approx4.5 \\  \\ \sum(x-\bar{x})^2=30.41 \\  \\ s.d= \sqrt{ \frac{\sum(x-\bar{x})^2}{n} } = \sqrt{ \frac{30.41}{20} } = \sqrt{1.5205} =1.23

Therefore, the standard deviation of x-bar is approximately 1.23.
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Jaina and Tomas compare their compound interest accounts to see how much they will have in the accounts after three years. They
MariettaO [177]

The question is incomplete. The complete question is :

Jaina and Tomas compare their compound interest accounts to see how much they will have in the accounts after three years. They substitute their values shown below into the compound interest formula. Compound Interest Accounts Name Principal Interest Rate Number of Years Compounded Jaina $300 7% 3 Once a year Tomas $400 4% 3 Once a year. Which pair of equations would correctly calculate their compound interests?

Solution :

It is given that Jaina and Tomas wants to open an account by depositing a principal amount for a period of 3 years and wanted to calculate the amount they will have using the compound interest formula.

<u>So for Jiana</u> :

Principal, P = $300

Rate of interest, r = 7%

Time, t = 3

Compounded yearly

Therefore, using compound interest formula, we get

$A=P\left(1+\frac{r}{100}\right)^{t}$

   $=300\left(1+\frac{7}{100}\right)^{3}$

   $=300(1+0.07)^3$

<u>Now for Tomas </u>:

Principal, P = $400

Rate of interest, r = 4%

Time, t = 3

Compounded yearly

Therefore, using compound interest formula, we get

$A=P\left(1+\frac{r}{100}\right)^{t}$

   $=400\left(1+\frac{4}{100}\right)^{3}$

   $=400(1+0.04)^3$

Therefore, the pair of equations that would correctly calculate the compound interests for Jaina is $A=300(1+0.07)^3$ .

And the pair of equations that would correctly calculate the compound interests for Tomas is $A=400(1+0.04)^3$ .

8 0
2 years ago
Read 2 more answers
One week, Claire earned $272.00 at her job when she worked for 17 hours. If she is paid the same hourly wage, how many hours wou
DochEvi [55]

Answer: 448 hours

Step-by-step explanation:

272 divided by 17 = hourly wage

hourly wage= 16$

448 divided by 16 = 28 weeks

28 x 16= 448 hours

4 0
2 years ago
Read 2 more answers
This background applies to the next several questions. Assume a 15 cm diameter wafer has a cost of 12, contains 84 dies, and has
nalin [4]

Answer:

1) Yield_1= \frac{1}{(1+ 0.02 \frac{1}{2} 2.104)^2}=0.959

Yield_2= \frac{1}{(1+ 0.031 \frac{1}{2} 3.1415)^2}=0.909

2) Cost/die_1 = \frac{12}{84 x 0.959}=0.149

Cost/die_2 = \frac{15}{100 x 0.909}=0.165

3) Area_1 = \frac{1.1 \pi (7.5cm)^2}{84}=\frac{2.104 cm^2}{1.1}=1.913 cm^2

Area_2 = \frac{1.1 \pi (10cm)^2}{100}=\frac{3.1415 cm^2}{1.1}=2.856 cm^2

And for the new yield we need to take in count the increase of 15% for the area and we got this:

Yield_1= \frac{1}{(1+(1.15) 0.02 \frac{1}{2} 1.913)^2}=0.957

Yield_2= \frac{1}{(1+(1.15) 0.031 \frac{1}{2} 2.856)^2}=0.905

4) DR_{old}=\frac{1}{\sqrt{0.92}} -1=0.0426 defects/cm^2

DR_{new}=\frac{1}{\sqrt{0.95}} -1=0.0260defects/cm^2

Step-by-step explanation:

Part 1

For this part first we need to find the die areas with the following formula:

Area= \frac{W area}{Number count}

Area_1 = \frac{\pi (7.5cm)^2}{84}=2.104 cm^2

Area_2 = \frac{\pi (10cm)^2}{100}=3.1415 cm^2

Now we can use the yield equation given by:

Yield=\frac{1}{(1+ DR\frac{Area}{2})^2}

And replacing we got:

Yield_1= \frac{1}{(1+ 0.02 \frac{1}{2} 2.104)^2}=0.959

Yield_2= \frac{1}{(1+ 0.031 \frac{1}{2} 3.1415)^2}=0.909

Part 2

For this part we can use the formula for cost per die like this:

Cost/die = \frac{Cost per day_i}{Number count_i x Yield_i}

And replacing we got:

Cost/die_1 = \frac{12}{84 x 0.959}=0.149

Cost/die_2 = \frac{15}{100 x 0.909}=0.165

Part 3

For this case we just need to calculate the new area and the new yield with the same formulas for part a, adn we got:

Area_1 = \frac{1.1 \pi (7.5cm)^2}{84}=\frac{2.104 cm^2}{1.1}=1.913 cm^2

Area_2 = \frac{1.1 \pi (10cm)^2}{100}=\frac{3.1415 cm^2}{1.1}=2.856 cm^2

And for the new yield we need to take in count the increase of 15% for the area and we got this:

Yield_1= \frac{1}{(1+(1.15) 0.02 \frac{1}{2} 1.913)^2}=0.957

Yield_2= \frac{1}{(1+(1.15) 0.031 \frac{1}{2} 2.856)^2}=0.905

Part 4

First we can convert the area to cm^2 and we got 2 cm^2 the yield would be on this case given by:

Yield= \frac{1}{(1+DR\frac{2cm^2}{2})^2}=\frac{1}{1+(DR)^2}

And if we solve for the Defect rate we got:

DR= \frac{1}{\sqrt{Yield}}-1

Now we can find the previous and new defect rate like this:

DR_{old}=\frac{1}{\sqrt{0.92}} -1=0.0426 defects/cm^2

And for the new defect rate we got:

DR_{new}=\frac{1}{\sqrt{0.95}} -1=0.0260defects/cm^2

5 0
2 years ago
A bicyclist travels at a constant speed of 12 miles per hour for a total of 45 minutes. (Use set notation for the domain and ran
Mamont248 [21]

A bicyclist travels at a constant speed of 12 miles per hour for a total of 45 minutes.

We know the formula , Distance = speed * time

Speed is constant and it is 12. So it  is linear

The function becomes d = 12t, x is the t is the time and d is the distance

At the starting point, t=0  and distance d=0

End point , t=45 min = 0.75 hours and distance = 12 * 0.75 = 9

So domain (t) is {x|0}

Range (d) is {y|0}

5 0
2 years ago
Read 2 more answers
The area of a room is 396 square feet. The length is x+3, and the width is x+7 feet. Find the dimensions of the room
Marrrta [24]
This is an odd problem. The length and width look like they should be interchanged. Anyway I'll solve it and we'll talk about the results.
(x +7) (x + 3) = 396
x^2 + 7x + 3x + 21 = 396
x^2 + 10x + 21 = 396 Subtract 21 from both sides.
x^2 + 10x = 396 - 21
x^2 + 10x = 375
x^2 + 10x - 375 = 0 This probably factors.
(x + 25)(x - 15)
x + 25 = 0
x = - 25 which makes no sense. Negatives do not describe room dimensions.
x - 15= 0
x = 15. this is fine.

x+3 = 15 + 3 = 18
x + 7 = 15 + 7= 22

Check
=====
Area = L * W = 18 * 22 = 496. It does check.
22 should be the length

18 should be the width.
3 0
2 years ago
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