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madam [21]
3 years ago
11

Resistors for electronic circuits are manufactured on a high-speed automated machine. The machine is set up to produce a large r

un of resistors of 1,000 ohms each. Use Exhibit 10.13. To set up the machine and to create a control chart to be used throughout the run, 15 samples were taken with four resistors in each sample. The complete list of samples and their measured values are as follows: Use three-sigma control limits. SAMPLE NUMBER READINGS (IN OHMS) 1 1027 994 977 994 2 975 1013 999 1017 3 988 1016 974 997 4 998 1024 1006 1010 5 990 1012 990 1000 6 1016 998 1001 1030 7 1000 983 979 971 8 973 982 975 1030 9 992 1028 991 998 10 997 1026 972 1021 11 990 1021 1028 992 12 1021 998 996 970 13 1027 993 996 996 14 1022 981 1014 983 15 977 993 986 983 a. Calculate the mean and range for the above samples.
Mathematics
1 answer:
Anettt [7]3 years ago
8 0

Answer:

See explanation

Step-by-step explanation:

Given

See attachment for proper presentation of question

Required

Mean and Range

To do this, we simply calculate the mean and the range of each row.

\bar x = \frac{\sum x}{n} ---- mean

Where:

n = 4 ---- number of rows

R = Highest - Lowest --- range

So, we have:

Sample 1

\bar x_1 = \frac{1027+ 994 +977 +994 }{4}

\bar x_1 = 998

R_1 = 1027- 994

R_1 = 33

Sample 2

\bar x_2 = \frac{975 +1013 +999 +1017}{4}

\bar x_2 = 1001

R_2 =  1017 - 975

R_2 = 42

Sample 3

\bar x_3 = \frac{988 +1016 +974 +997}{4}

\bar x_3 = 993.75

R_3 = 1016-974

R_3 = 42

Sample 4

\bar x_4 = \frac{998 +1024 +1006 +1010}{4}

\bar x_4 = 1009.5

R_4 = 1024 -998

R_4 = 26

Sample 5

\bar x_5 = \frac{990 +1012 +990 +1000}{4}

\bar x_5 = 998

R_5 = 1012 -990

R_5 = 22

Sample 6

\bar x_6= \frac{1016 + 998 +1001 +1030}{4}

\bar x_6= 1011.25

R_6= 1030-998

R_6= 32

Sample 7

\bar x_7 = \frac{1000 +983 +979 +971}{4}

\bar x_7 = 983.25

R_7 = 1000-971

R_7 = 29

Sample 8

\bar x_8 = \frac{973 +982 +975 +1030}{4}

\bar x_8 = 990

R_8 = 1030-973

R_8 = 57

Sample 9

\bar x_9 = \frac{992 +1028 +991 +998}{4}

\bar x_9 = 1002.25

R_9 = 1028 -991

R_9 = 37

Sample 10

\bar x_{10} = \frac{997 +1026 +972 +1021}{4}

\bar x_{10} = 1004

R_{10} = 1026 -972

R_{10} = 54

Sample 11

\bar x_{11} = \frac{990 +1021 +1028 +992}{4}

\bar x_{11} = 1007.75

R_{11} = 1028 -990

R_{11} = 38

Sample 12

\bar x_{12} = \frac{1021 +998 +996 +970}{4}

\bar x_{12} = 996.25

R_{12} = 1021 -970

R_{12} = 51

Sample 13

\bar x_{13} = \frac{1027 +993 +996 +996}{4}

\bar x_{13} = 1003

R_{13} =1027 -993

R_{13} =34

Sample 14

\bar x_{14} = \frac{1022 +981 +1014 +983}{4}

\bar x_{14} = 1000

R_{14} = 1022 -981

R_{14} = 41

Sample 15

\bar x_{15} = \frac{977 +993 +986 +983}{4}

\bar x_{15} = 984.75

R_{15} = 993-977

R_{15} = 16

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Meg lives in Indianapolis and wants to visit her mom in Lima. She has been meaning to go to a chiropractor in Dayton, so she is
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Answer: 44 miles

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The distance between Indianapolis and Dayton, ID = 165 miles
The distance between Dayton and Lima, DL is unknown

Since there are straight roads connecting the three cities, the connection between them form a right angles triangle.

The right angle is at Dayton
The hypotenuse is the distance between Indianapolis and Lima, IL

Therefore IL^2 = ID^2 + DL^2
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An angelfish was 1 1/2 inches long when it was bought. Now it is 2 1/3 inches long.
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A)

Earlier, The length of the angelfish  = 1 \frac{1}{2} inches

Now, the length of angelfish = 2 \frac{1}{3} inches

We have to determine the grown length of angelfish

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= \frac{7}{3}- \frac{3}{2}

LCM of '2' and '3' is '6',

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= \frac{5}{6} inch

Therefore, the angelfish has grown by \frac{5}{6} inch.

B)

We have to determine the increased length of angelfish in feet.

Since 1 inch = \frac{1}{12} foot

So, \frac{5}{6} inch = \frac{5}{6} \times \frac{1}{12} = \frac{5}{72}

= 0.069 foot.

7 0
3 years ago
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