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### Solve for [tex]\( x \)[/tex].
[tex]\[ 3x = 6x - 2 \][/tex]

### Types of Variation
Question 9, 6.B.29
HW Score: 28.7\%, 6.6 of 23 points

Part 3 of 6
Points: 0 of 4

An auto transmission manufacturer receives ball bearings from two different suppliers. The ball bearings must have a specified diameter of 16.30 mm with a tolerance of [tex]\(\pm 0.1 \)[/tex] mm. Recent shipments from the two suppliers had ball bearings with the following diameters. Complete parts (a) through (c).

[tex]\[
\begin{tabular}{llllllll}
Supplier A: & 16.23 & 16.27 & 16.32 & 16.33 & 16.36 & 16.42 & 16.45 \\
Supplier B: & 16.18 & 16.21 & 16.25 & 16.32 & 16.39 & 16.43 & 16.45
\end{tabular}
\][/tex]

[tex]\[
\begin{array}{l}
\text{mean} = 16.32 \\
s = 0.11
\end{array}
\][/tex]
(Round to the nearest hundredth as needed.)

b. Draw a boxplot for each data set, and mark the tolerance on each boxplot.

Draw a boxplot for the diameters of the ball bearings from Supplier A. Choose the correct answer below.

A.
B.
C.
D.


Sagot :

Sure, let's go through the solution step-by-step.

### Part (a): Calculating Mean and Standard Deviation
To solve this question, we will start by ensuring we have the basic statistical measures calculated for each sample set: mean and standard deviation. For both sets, Supplier A and Supplier B.

#### Supplier A Data
Diameters: 16.23, 16.27, 16.32, 16.33, 16.36, 16.42, 16.45

#### Supplier B Data
Diameters: 16.18, 16.21, 16.25, 16.32, 16.39, 16.43, 16.45

Mean Calculation:

The mean [tex]\( \mu \)[/tex] is calculated as:
[tex]\[ \mu = \frac{\sum x_i}{n} \][/tex]
Where [tex]\( x_i \)[/tex] are the measurements and [tex]\( n \)[/tex] is the number of measurements.
For Supplier A:
[tex]\[ \mu_A = \frac{16.23 + 16.27 + 16.32 + 16.33 + 16.36 + 16.42 + 16.45}{7} = \frac{114.38}{7} \approx 16.34 \][/tex]

For Supplier B:
[tex]\[ \mu_B = \frac{16.18 + 16.21 + 16.25 + 16.32 + 16.39 + 16.43 + 16.45}{7} = \frac{114.23}{7} \approx 16.32 \][/tex]

Standard Deviation Calculation:

The standard deviation [tex]\( s \)[/tex] is calculated as:
[tex]\[ s = \sqrt{\frac{\sum (x_i - \mu)^2}{n-1}} \][/tex]

### Part (b): Drawing the Boxplot
To draw the boxplot, we need the five-number summary: minimum, first quartile (Q1), median (Q2), third quartile (Q3), and maximum.

#### Supplier A:
Sorted Data: 16.23, 16.27, 16.32, 16.33, 16.36, 16.42, 16.45

1. Minimum (Min): 16.23
2. First Quartile (Q1): median of the first half of the data
- Data below 16.33: 16.23, 16.27, 16.32
- Q1 = median of (16.23, 16.27, 16.32) = 16.27
3. Median (Q2): 16.33
4. Third Quartile (Q3): median of the second half of the data
- Data above 16.33: 16.36, 16.42, 16.45
- Q3 = median of (16.36, 16.42, 16.45) = 16.42
5. Maximum (Max): 16.45

#### Supplier B:
Sorted Data: 16.18, 16.21, 16.25, 16.32, 16.39, 16.43, 16.45

1. Minimum (Min): 16.18
2. First Quartile (Q1): 16.21
3. Median (Q2): 16.32
4. Third Quartile (Q3): 16.43
5. Maximum (Max): 16.45

The tolerance levels for diameters are:
- Lower Tolerance: 16.30 - 0.10 = 16.20 mm
- Upper Tolerance: 16.30 + 0.10 = 16.40 mm

### Boxplot Drawing
To draw the boxplots with the tolerances marked:
1. Create the box for each supplier using the five-number summary.
2. Add the tolerance lines at 16.20 mm and 16.40 mm.

### Boxplot for Supplier A:
- Min: 16.23
- Q1: 16.27
- Median: 16.33
- Q3: 16.42
- Max: 16.45

### Boxplot for Supplier B:
- Min: 16.18
- Q1: 16.21
- Median: 16.32
- Q3: 16.43
- Max: 16.45

The boxplots should look like this:

#### Supplier A:
```
A. _______|______|_________|_________
16.20 16.27 16.33 16.42 16.40
```

#### Supplier B:
```
B. ______|______|________|_________
16.18 16.21 16.32 16.43 16.40
```

Ensure that the tolerance lines are clearly marked and specific to the required values of 16.20 mm and 16.40 mm in the plot.

The correct answer is based on visually matching this description to the answer options provided in your question.
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