Answer :
Answer:
Sample mean = 570.5
Median = 571
Range = 10
Variance = 10
Standard deviation = 3.162
Step-by-step explanation:
We are given the following in the question:
572, 572, 573, 568, 569, 575, 565, 570
Sorted data: 565, 568, 569, 570, 572, 572, 573, 575
a) Sample mean
Formula:
[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]
[tex]Mean =\displaystyle\frac{4564}{8} = 570.5[/tex]
[tex]Median:\\\text{If n is odd, then}\\\\Median = \displaystyle\frac{n+1}{2}th ~term \\\\\text{If n is even, then}\\\\Median = \displaystyle\frac{\frac{n}{2}th~term + (\frac{n}{2}+1)th~term}{2}[/tex]
[tex]\text{Median} = \displaystyle\frac{4^{th}+5^{th}}{2} = \frac{570+572}{2} = 571[/tex]
b) Range = Maximum - Minimum = 575 - 565 = 10
Formula:
[tex]\text{Variance} = \displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}[/tex]
where [tex]x_i[/tex] are data points, [tex]\bar{x}[/tex] is the mean and n is the number of observations.
Sum of squares of differences = 70
[tex]\sigma^2= \sqrt{\dfrac{70}{7}} = 10\\\\\sigma = \sqrt{70} = 3.162[/tex]
c) The mean diameter is approximately same as the ideal diameter. The diameter lies within 3.162 mm of the mean diameter. Thus, the quality of tires is good.
For the considered sample of the tire diameters' data, we get:
- Sample mean: 570.5
- Median of the sample: 571
- sample variance: 10
- Standard deviation: 3.162 approx.
- Range of the sample: 10
- Quality of tires: Almost ideal as the mean is close to ideal 570 mm diameter, and standard deviation is 3.16 which in comparison to scale of 500 approx number is quite low).
How to find mean and median of a data?
Mean is the ratio of sum of the observations to the total number of observations.
Median is such a number for the arranged data set(ascending or descending order) such that to its left and to its right belong same number of observations.
How to find the variance and standard deviation of a data set?
Suppose that [tex]x_i; \: \: i = 1,2, ... ,n[/tex] are n data values for the given data set.
Then we have:
Variance =[tex]\sigma^2 = \dfrac{\sum_{\forall x_i} (x_i - \overline{x})^2}{n}[/tex]
As standard deviation is positive root of variance, thus,
[tex]\sigma = \sqrt {\dfrac{\sum_{\forall x_i} (x_i - \overline{x})^2}{n}}[/tex]
If its sample, we divide by n-1 instead of n
The range of a data set is the difference between its maximum and minimum value.
For this case, we have the diameter of the tires as:
572, 572, 573, 568, 569, 575, 565, 570
They are total n = 8 values.
Arranging the data in ascending order, we get:
565, 568, 569, 570, 572, 572, 573, 575
The maximum value = 575
The minimum value = 565
Thus, range of the data set = Max - min = 575 - 565 = 10
The central value pair for 8 values is fourth and fifth value. Any value in between fourth and fifth value can be considered median as on its left and right will lie same amount of observations, but as tradition goes, we take the average. Thus, we get:
median of the data set = [tex]\dfrac{570 + 572}{2} = 571[/tex]
The sum of all 8 values is 4564, thus, we get:
Mean of the data = sum of all values/count of values = 4564/8 = 570.5
Thus, [tex]\overline{x} = 570.5[/tex]
Using the formula for variance, but since it is sample, we divide by n-1 instead of n, we get:
[tex]\sigma^2 = \dfrac{\sum_{\forall x_i} (x_i - \overline{x})^2}{n-1} = \dfrac{\sum_{\forall x_i} (x_i - 570.5)^2}{7} \\\\\sigma^2 = \dfrac{2.25+ 2.25+ 6.25 + 6.25+ 2.25+ 20.25 +30.25+ 0.25}{8} = 10[/tex]
Thus, standard deviation of this data set is [tex]\sigma = \sqrt{10} \approx 3.162[/tex]
Thus, for the considered sample of the tire diameters' data, we get:
- Sample mean: 570.5
- Median of the sample: 571
- sample variance: 10
- Standard deviation: 3.162 approx.
- Range of the sample: 10
- Quality of tires: Almost ideal as the mean is close to ideal 570 mm diameter, and standard deviation is 3.16 which in comparison to scale of 500 approx number is quite low).
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