A manufacturer of band saws wants to estimate the average repair cost per month for the saws he has sold to certain industries. He cannot obtain a repair cost for each saw, but he can obtain the total amount spent for saw repairs and the number of saws owned by each industry. Thus, he decides to use cluster sampling, with each industry as a cluster. The manufacturer selects a simple random sample of n = 20 from the N = 96 industries he services. The data on total cost of repairs per industry and number of saws per industry are as given in the accompanying table. Estimate the average repair cost per saw for the past month and place a bound on the error of estimation.
Industry No.of saws Total repair cost for past month
1 3 50
2 7 110
3 11 230
4 9 140
5 2 60
6 12 280
7 14 240
8 3 45
9 5 60
10 9 230
11 8 140
12 6 130
13 3 70
14 2 50
15 1 10
16 4 60
17 12 280
18 6 150
19 5 110
20 8 120

Answers

Answer 1

Answer:

The average repair cost per saw for the past month is 20.

The bound of error of estimation is ±10.

Step-by-step explanation:

a) Data and Calculations:

Industry        No.of saws   Total repair cost   Average

                                           for past month

1                            3                        50           16.67

2                           7                        110           15.71

3                          11                       230          20.91

4                          9                        140           15.56

5                          2                         60           30

6                         12                       280           23.33

7                         14                       240           17.14

8                          3                         45           15

9                          5                        60           12

10                         9                      230           25.56

11                          8                       140           17.50

12                         6                       130           21.67

13                         3                        70           23.33

14                         2                        50           25

15                         1                          10           10

16                        4                         60           15

17                       12                       280           23.33

18                        6                        150           25

19                        5                         110           22

20                       8                        120           15

Total               130                    2,565           19.73

Average = Sum of the total repair cost for past month divided by the number of saws repaired

= 19.73 (2,565/130)

= 20

The bound on error of estimation = the difference between the upper bound of the interval and the calculated mean

= 30 - 20

= 10

The lower bound = 10

The bound of error is also = (30 -10)/2 = 10


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Answers

Answer:

Answer: 50∘ and 130∘

Step-by-step explanation:

deltamath:)

The measures of M are 50.0 and 130 degrees

The given parameters are:

[tex]\mathbf{m = 5.6cm}[/tex]

[tex]\mathbf{o = 4.6cm}[/tex]

[tex]\mathbf{\angle O = 39^o}[/tex]

The measure of angle M is calculated using the following sine formula

[tex]\mathbf{\frac{o}{sin\ O} = \frac{m}{sin\ M}}[/tex]

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[tex]\mathbf{M = 180 -50.0}[/tex]

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Answers

Answer:

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Step-by-step explanation:

Answer: (d)

Step-by-step explanation:

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Answers

Answer:

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Step-by-step explanation:

delta math

The required value of  ∠J is 56.6° for the given triangle.

What is a Triangle?

Triangle is defined as a basic polygonal shape of a triangle that has three sides and three interior angles. It is one of the fundamental shapes in geometry and is represented by the symbolΔ.

Here, we are given:

j = 6.3 inches (the length of side JK)

l = 6.1 inches (the length of side KL)

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We want to find the possible values of ∠J (the measure of angle J).

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∠J = 56.6°

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Answer:

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Answers

Answer:

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Answer:

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The organizers would call this event a success if the total contributions exceed $4,600

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Mean and a standard deviation of $105 and $40

This means that [tex]\mu = 105, \sigma = 40[/tex]

Sample of 40

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What is the probability that this fund-raising event is a success?

Sample mean above 115, which is 1 subtracted by the pvalue of Z when X = 115. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

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[tex]Z = 1.58[/tex]

[tex]Z = 1.58[/tex] has a pvalue of 0.9429

1 - 0.9429 = 0.0571

0.0571 = 5.71% probability that this fund-raising event is a success

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Answers

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Step-by-step explanation:

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Sample Information:

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Alternative Hypothesis     Hₐ                 x ≠  μ

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Answers

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9514 1404 393

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Step-by-step explanation:

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