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The operations manager of a plant that manufactures tires wishes to compare the actual inner diameter of two grades of tires, each of which has a nominal value of 575 millimeters. A sample of five tires of each grade is selected, and the results representing the inner diameters of the tires, ordered from smallest to largest, are as follows:
Grade X: 568, 570, 575, 578, 584
Grade Y: 573, 574, 575, 577, 578
a. Compute the mean, median, and standard deviation for each grade of tire.
b. What would be the effect on your answers in (a) and (b) if the last value for grade Y was 588 instead of 578? Explain.
c. Which grade of tire is providing better quality? Explain.

Answer :

MrRoyal

Answer:

See Explanation for answers

Step-by-step explanation:

Given

[tex]Grade\ X: 568, 570, 575, 578, 584[/tex]

[tex]Grade\ Y: 573, 574, 575, 577, 578[/tex]

Solving (a): Mean, median, and standard deviation of both grades

Mean is calculated as:

[tex]Mean = \frac{\sum x}{n}[/tex]

For grade X:

[tex]Mean = \frac{568 + 570+ 575+578+584}{5}[/tex]

[tex]Mean = \frac{2875}{5}[/tex]

[tex]Mean = 575[/tex]

For Grade Y

[tex]Mean = \frac{573+ 574+ 575+ 577+ 578}{5}[/tex]

[tex]Mean = \frac{2877}{5}[/tex]

[tex]Mean = 575.4[/tex]

[tex]Median = \frac{N+1}{2}th[/tex]

[tex]Median = \frac{5+1}{2}th[/tex]

[tex]Median = \frac{6}{2}th[/tex]

[tex]Median = 3rd\ item[/tex]

For grade X and Y.

[tex]Median = 575[/tex]

Standard deviation (s) is calculated as:

[tex]s = \sqrt{\frac{\sum(x_i - \bar x)^2}{n}}[/tex]

For Grade x

[tex]s = \sqrt{\frac{(568 - 575)^2+(570 - 575)^2+(575 - 575)^2+(578- 575)^2+(584- 575)^2}{5}}[/tex]

[tex]s = \sqrt{\frac{164}{5}}[/tex]

[tex]s = \sqrt{32.8}[/tex]

[tex]s = 5.73[/tex]

For Grade Y

[tex]s = \sqrt{\frac{(573 - 575.4)^2+(574 - 575.4)^2+(575 - 575.4)^2+(577 - 575.4)^2+(578 - 575.4)^2}{5}}[/tex]

[tex]s = \sqrt{3.44}[/tex]

[tex]s = 1.85[/tex]

Solving (b): If the last value is  b, the following will occur.

The mean will increase.

The median will remain unaltered

The standard deviation will increase.

See proof below

For Grade Y: Mean is:

[tex]Mean = \frac{573+ 574+ 575+ 577+ 588}{5}[/tex]

[tex]Mean = \frac{2887}{5}[/tex]

[tex]Mean = 577.4[/tex]

Median still remains the 3rd item.

[tex]Median = 575[/tex]

The standard deviation is:

[tex]s = \sqrt{\frac{(573 - 575.4)^2+(574 - 575.4)^2+(575 - 575.4)^2+(577 - 575.4)^2+(588 - 575.4)^2}{5}}[/tex]

[tex]s = \sqrt{\frac{169.2}{5}}[/tex]

[tex]s = \sqrt{33.84}[/tex]

[tex]s = 5.82[/tex]

Solving (c): Which grade provides better quality.

Grade X provides better quality.

This is so because the mean value, the standard deviation of grade X is greater than grade Y

Mean medium and standard deviation are the statistics tools used to measure the center tendency of the given data set.

  • The mean, medium standard deviation are computed and value obtained.
  •  If value last value of data set Y is 588 instead of 578 the mean and standard deviation will increase but medium will remain same for data set Y.
  • The value of mean and standard deviation of the Tyre X is greater thus the Tyre X provide the better quality.

Given information-

Grade X: 568, 570, 575, 578, 584

Grade Y: 573, 574, 575, 577, 578

  • a) Compute the mean, median, and standard deviation for each grade of tire.

Mean

Mean is the average of the given data or the ratio of the sum of the given data set to the number of data in the set.

Mean for grade X,

[tex]\overline x=\dfrac{568+570+575+578+584}{5} \\ \overline x=575[/tex]

Mean for grade Y,

[tex]\overline x=\dfrac{573+574+575+577+578}{5} \\ \overline x=575.4[/tex]

Median

Median is the middle number when the data set is arranged in the assenting or descending order.

The middle number for both the data set is 575 milimeter. Thus the median is 575 for both the data set.

Standard deviation

Standard deviation refers that how much the group member of a data set is differ from the mean value. Standard deviation can be given as,

[tex]\sigma=\sqrt{\dfrac{\sum(x_i-\mu)^2}{N} } [/tex]

Standard deviation for X

[tex]\sigma=\sqrt{\dfrac{(568-575)^2+(570-575)^2+(575-575)^2+(578-575)^2+(584-575)^2}{5} } \\ \sigma=5.73[/tex]

Standard deviation for Y

[tex]\sigma=\sqrt{\dfrac{(573-575.4)^2+(574-575.4)^2+(575-575.4)^2+(577-575.4)^2+(578-575.4)^2}{5} } \\\sigma=1.85[/tex]

b) Effect on the answers in (a) and (b) if the last value for grade Y was 588 instead of 578.

Mean for grade Y,

[tex]\overline x=\dfrac{573+574+575+577+588}{5} \\ \overline x=577.4[/tex]

Thus the mean increased with value 2 mm.

Medium remain the same as the middle value is still 575.

Standard deviation for Y

[tex]\sigma=\sqrt{\dfrac{(573-575.4)^2+(574-575.4)^2+(575-575.4)^2+(577-575.4)^2+(588-575.4)^2}{5} } \\\sigma=5.82[/tex]

c) Tire which provide better quality-

The value of mean and standard deviation of the Tyre X is greater thus the Tyre X provide the better quality.

Hence

  • The Mean medium and standard deviation are computed and value obtained.
  •  If value last value of data set Y is 588 instead of 578 the mean and standard deviation will increase but medium will remain same for data set Y.
  • The value of mean and standard deviation of the Tyre X is greater thus the Tyre X provide the better quality.

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